答:
The definition of the identity element is $$\exists e, \forall x, e*x=x*e=x$$

and not
$$\forall x, \exists e, e*x=x*e=x$$

Title: Two identity element?

Body:
On $\Bbb N=\{0,1,2,...\}$ we define the operation $\otimes$ by $m\otimes n= |m-n|$. Are there any identity element?

I came up with two identity elements. $e= 0$ and $e=2m$ for an element $m$ in the set, but how is that possible? I thought on a operation $\otimes$ that it would always be one identity element?


问:
Proving that this integral exists both as Riemann Integral and Lebesgue integral
Q 10.9  Apostol Mathematical analysis

10.9 (a) If p>1, prove that the integral $\int_{1}^{+\infty} x^{-p} sin x  dx$ exists  both as improper riemann integral and lebesgue integral.

For proving riemann integral I have to prove that for all b$\geq$ a Lim $ b\to \infty \int_{a}^{b} f(x) dx$ exists and for proving lebesgue integral I need to prove that there exists a positive integer M such that $\int_{b}^{a}|f(x)| d x \leq M$ for every b$\geq a$.

But I am not able to prove any if the results.

For !st I used integration by parts 1 time and in integral in 2nd term  I got power as $1/x^{p+1} $ , so I stopped as power would further increase .

But If i  use improper integral test with $1/x^{1+\delta}$ ,$\delta >0$ we will get that integral is convergent as $\int_{1}^{\infty}1/x^{1+\delta} dx$  .

But I am not sure how to show the condition for existence of lebesgue integral.

Can you please help with that

Edit: I have a question suppose that $\int_{1}^{\infty}x^{-p} sin x dx $ and 0<p$\leq 1$ then How to prove that The integral exists as improper integral . The method used above can't be usedas p<1 so I am not able to proceed with proving why it is reimann integrable.

Ty

答:
Using integration by parts is a fine approach and, in fact, shows that the improper Riemann integral converges for $p > 0$.  This is also a consequence of the Dirichlet test.

The question only considers $p > 1$, and showing that the improper Riemann integral  converges in this case is even easier.   We have  $|x^{-p} \sin x | \leqslant x^{-p}$ and $\int_1^\infty x^{-p} \, dx = = 1/(p-1) < \infty$ when $p > 1$. Hence, the improper Riemann integral converges as a consequence of the Weierstrass M-test.

For Lebesgue integrability, with $f(x) = x^{-p} \sin x$ and since the Riemann and Lebesgue integrals coincide on finite intervals, we have

$$\int_{[1,\infty]}|f|\, \chi_{[1,b]}= \int_{[1,b]}|f| = \int_1^b |f(x)| \, dx $$

Since $f$ is nonnegative, it follows from the monotone convergence theorem and the convergence of the improper integral of $|f|$ that

$$\int_{[1,\infty]}|f|= \int_{[1,\infty]}\lim_{b \to \infty}|f|\, \chi_{[1,b]}= \lim_{b \to \infty} \int_{[1,\infty]}|f|\, \chi_{[1,b]} = \lim_{b \to \infty}\int_1^b |f(x)| \, dx < \infty$$

ContentLicense: CC BY-SA 4.0


问:
How to prove uniqueness of approximate "jump" values of an $L^1$ function?
Let $v \in \mathbb{S}^{n-1}$ be a unit vector. Given $r>0$, Let $B_r(x)$ be the ball of radius $r$ around $x \in \mathbb{R}^n$, and define

$$B_r^+(x,v)=\{ y \in B_r(x) \, | \, \langle y-x,v\rangle >0\},$$
and
$$B_r^-(x,v)=\{ y \in B_r(x) \, | \, \langle y-x,v\rangle <0 \}. $$

Now, let $\Omega \subseteq \mathbb{R}^n$ be an open subset, and let $u \in L^1_{\text{loc}}(\Omega)$.

Let $a \neq b \in \mathbb{R}$, and suppose that

$$
\lim_{r \to 0} \frac{1}{\mu(B_r^+(x,v))}\int_{B_r^+(x,v)}  |u(y)-a|dy=0,
$$
and
$$
\lim_{r \to 0} \frac{1}{\mu(B_r^-(x,v))}\int_{B_r^-(x,v)}  |u(y)-b|dy=0.
$$

Question: How to prove that the triple $(a,b,v)$ is uniquely determined up to a permutation of $a$ and $b$, and a change of sign of $v$?

(I stumbled on this fact in a geometric measure theory book- it seems rather plausible, but I don't see an immediate way to prove it).

答:
Up to translation and rescaling, this is a claim about a sequence of functions in $B(0,1)$.  To wit, define $u_{r} : B(0,1) \to \mathbb{R}$ by $u_{r}(\xi) = u(x + r \xi)$ to find that
\begin{equation*}
\lim_{r \to 0^{+}} \int_{B^{+}_{1}(0,v)} |u_{r}(\xi) - a| \, d \xi = 0, \quad \lim_{r \to 0^{-}} \int_{B^{-}_{1}(0,v)} |u_{r}(\xi) - b| \, d \xi = 0.
\end{equation*}
Now that the domain is no longer moving, this is a question about $L^{1}$-convergence of functions in $B(0,1)$.  In fact, notice that
\begin{equation*}
\lim_{r \to 0^{+}} \int_{B_{1}(0)} |u_{r}(\xi) - a \chi_{B_{1}^{+}(0,v)} - b \chi_{B_{1}^{+}(0,v)}| \, d\xi = 0.
\end{equation*}
Since limits in $L^{1}(B(0,1))$ are unique and $a \chi_{B_{1}^{+}(0,v)} + b \chi_{B_{1}^{+}(0,v)}$ uniquely determines $a$, $b$, and $v$ (up to the permutation $(a,b,v) \mapsto (b,a,-v)$), we are done.


问:
Showing two measures are equal up to scalar multiplication.
I've been stuck on the following question and was wondering if it'd be possible to get some help.

Let $\mu, \nu$ be two finite Borel measures on a metric space $X$ s.t. $\mu << \nu$ and let $c>0$ and suppose that $\nu$ is doubling and that $$\lim_{r \to 0} \frac{\mu(B_r(x))}{\nu(B_r(x))} = c$$ for $\nu$ almost every $x\in X$. By using the Vitali Covering theorem to show that $\mu = c\nu$. (We assume the balls are closed)

I've been able to prove this using Radon-Nikodym combined with the lebesgue differentiation theorem but wish to prove it solely using Vital's covering theorem but have been unsuccessful in even finding an appropriate Vitali cover.

Theorem(Vitali Covering Theorem)

Let X be a metric space and let $\nu$ be a doubling measure on $X$ and let $\mathcal{B}$ be a Vitali cover for $S \subset X$ then there exists a countable $\mathcal{B}' \subset \mathcal{B}$ such that all elements of $\mathcal{B}'$ are disjoint and $$\nu \bigg(S \setminus \bigcup_{B' \in \mathcal{B}'}B'\bigg)=0$$

Definition(Vitali Cover)

Let $S \subset X$ then $\mathcal{B}$ a collection of closed balls such that $\forall \epsilon >0 , \forall x \in S$ there exists $B \in \mathcal{B}$ such that $x \in B$ and $\rm{rad}(B) < \epsilon$ then we call $\mathcal{B}$ a Vitali cover for S.

Many Thanks.

答:
It helps to know that finite Borel measures on a metric space $X$ are regular. In particular, if $E \subset X$ is a Borel set, $\mu$ is a finite Borel measure, and $\epsilon > 0$, then there exists an open set $G \supset E$ with $\mu(G) < \mu(E) + \epsilon$.

Assume that the stated conditions on $\mu$ and $\nu$ hold (and in particular $\nu(B(x,r)) > 0$ for all $x \in X$ and $r > 0$).

Let $E \subset X$ be a Borel set. Fix a constant $t > 1$.

For every $x \in E$ there exists $r_x > 0$ with the property that $0 < r < r_x$ implies  $$\frac ct < \frac{\mu(B(x,r))}{\nu(B(x,r))} < ct,$$ which can be expressed as
$$0 < r < r_x \implies \frac ct \nu(B(x,r)) \le \mu(B(x,r)) \le ct \nu(B(x,r)).$$

Let $\epsilon > 0$ and let $G \supset E$ be an open set with $\nu(G) < \nu(E) + \epsilon$. Define a Vitali cover ${\cal V}$ of $E$ by
$$ {\cal V} = \{B(x,r) : x \in E,\ 0 < r < r_x,\ B(x,r) \subset G\}.$$  According to Vitali's covering theorem there exists a countable disjointed subfamily ${\cal V}'$ of $\cal V$ satisfying
$$\mu(E) \le \mu \left( E \setminus \bigcup_{B \in {\cal V}'} B \right) + \mu \left(\bigcup_{B \in {\cal V}'} B \right) \le \sum_{B \in {\cal V}'} \mu(B) \le ct\sum_{B \in {\cal V}'} \nu(B)$$
Since the family ${\cal V}'$ is pairwise disjoint and each $B \in {\cal V}'$ is a subset of $G$ you have in addition that
$$\sum_{B \in {\cal V}'} \nu(B) \le \nu(G) <\nu(E) + \epsilon.$$
Thus $\mu(E) < ct(\nu(E) + \epsilon)$ and by letting $\epsilon \to 0^+$ you get
$$\mu(E) \le ct \nu(E).$$

Now let $\epsilon > 0$ and let $H \supset E$ be an open set with $\mu(H) < \mu(E) + \epsilon$. Repeat the steps above to find that
$$\frac ct \nu(E) \le \mu(E)$$ and conclude
$$\frac ct \nu (E) \le \mu(E) \le ct \nu(E).$$ Finally let $t \to 1^+$.


问:
The sum of the integration of g and $g^{-1}$
Let $g$ be a strictly increasing continuous function mapping $[a,b]$ onto
$[A,B]$, and, as usual, let $g^{-1}: [A,B] \to [a,b]$ denote its inverse function.
Use geometric insight to visualize the equation $\int_a^b g + \int_A^B g^{-1} = bB - aA$.

Apply this to the function $g(\theta) = \tan \theta$ on the interval $[0,b]$ for a fixed $b \in (0,\pi/2)$.

Can someone give me some help on how to set this up?

答:
If you draw the graph of $g$ on the $x,y$-coordinate system, it contains exactly the $(x,g(x))$ points, $x\in [a,b]$. But these are the same as the $(g^{-1}(y),y)$ points, for $y\in [A,B]$.

So that, $\int_A^Bg^{-1}=\int_A^Bg^{-1}(y)\,dy$ appears on the picture horizontally, as the area from the $y$-axis to the graph of $g$.

Subtracting the two rectangles, $(\int_a^bg\,-(b-a)A)\ +\ (\int_A^Bg^{-1}\,-(B-A)a )$ we get the area of the rectangle $[a,b]\times [A,B]$.

ContentLicense: CC BY-SA 3.0


问:
Recursive function inequality
Let $x_1=1$ and $x_{n+1}=\tfrac{1}{1+x_n}$.
I want to proof that for every $n,m\in \mathbb{N}$ with $m\leq n$, the inequality $|x_{n}-x_{m}|\leq 2^{-m}$ is true.

This can be done by induction over $m$.
Two facts I found out are

$\tfrac{1}{2} \leq x_n \leq 1$

For every $n,m\in \mathbb{N}$ the inequality $|x_{n+1}-x_{m+1}|\leq \tfrac{1}{2}|x_n-x_m|$ is true.

Now the induction:
Begin: or $m=1$ we can see with 1. that it is true.
The Hypothesis is
$|x_{n}-x_{m+1}|\leq 2^{-(m+1)}$

With 2. we can see that
$ |x_{n+1}-x_{m+1}| \leq \tfrac{1}{2}|x_n-x_m| \leq \tfrac{1}{2} 2^{-m} = 2^{-(m+1)}$

But how to get from
$|x_{n}-x_{m+1}|$ to $|x_{n+1}-x_{m+1}|$?

Thanks in advance.

答:
I guess I solved it by myself.

$|x_{n}-x_{m+1}| \leq |x_{n+1}-x_{m+1}|$ because $x_n \leq x_{n+1}$ for every $x_n \in (-1, \tfrac{-1+\sqrt{5}}{2}]$ and as $\lim(x_n)=\tfrac{-1+\sqrt{5}}{2}$ for $n \rightarrow ∞$ the statement is true.

Title: Recursive function inequality


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

Body:
Fix $x\in[0,1]$. Using
$$ \sin(\frac12x)\ge \frac x2-\frac{1}{6}(\frac{x}{2})^3=\frac x2(1-\frac {x^2}{24})\ge \frac x2(1-\frac1{24})=\frac{23x}{48}  $$
you can have
\begin{eqnarray}
1-\cos x&=&2\sin^2(\frac x2)\ge2(\frac{23x}{48})^2>\frac{x^2}{3}.
\end{eqnarray}


问:
Finding the angle between $x-5=0$ and $2x+2y-7=0$
I have 2 line equations: $x-5=0$ and $2x+2y-7=0$

I believe the angle is $45^\circ$, but I don't know how to calculate the second equation when there's a $y$ as well.

Thanks!

答:
The standard form of equation for lines is
\begin{equation}
ax + by + c = 0,
\end{equation}
if you are talking about a 2D case.
The vector $\left(a,b\right)$ is the normal of the line. In your case, the normal for the first equation is $(1,0)$ while the normal for the latter is $\left(2,2\right)$. The dot product between the two normal vectors is
\begin{equation}
\left(1,0\right) \cdot \left(2,2\right) = 2.
\end{equation}
Also, we know that
\begin{equation}
a \cdot b = \lvert a\rvert \lvert b\rvert \cos{\theta},
\end{equation}
where $\theta$ is the angle between the two vectors. In your case, we have
\begin{equation}
2 = 1\cdot 2\sqrt{2} \cdot \cos{\theta},
\end{equation}
and this leads to
\begin{equation}
\cos{\theta} = \frac{1}{\sqrt{2}}.
\end{equation}
Thus, we have $\theta = 45^{\circ}$


问:
Logic problem - what kind of logic is it?
I would be most gratefull, if someone could verify my solution to this problem.

```<code> 1) Of Aaron, Brian and Colin, only one man is smart.

 Aaron says truthfully:
    1. If I am not smart, I will not pass Physics.
    2. If I am smart, I will pass Chemistry.

 Brian says truthfully:
    3. If I am not smart, I will not pass Chemistry.
    4. If I am smart, I will pass Physics.

 Colin says truthfully:
    5. If I am not smart, I will not pass Physics.
    6. If I am smart, I will pass Physics.

 While
   I. The smart man is the only man to pass one particular subject.
   II. The smart man is also the only man to fail the other particular subject.

 Which one of the three men is smart? Why?
</code>```

I would say that it could have been any one of them, as the implications in every statement are not strong enough to disprove the statements I,II. But I'm not sure if my solution is enough, as I'm not sure what kind of logic it is.

答:
Brian must be the smart man. For if Aaron is smart he passes Chemistry and so must be the only one to fail Physics, but Colin fails Physics. Likewise, if Colin is smart he passes Physics and must be the only one to fail Chemistry, but Brian fails Chemistry. Only Brian does not lead to a similar contradiction.

This is not much of a logic-based solution since simple elimination is enough here; I believe you can formalize it with standard propositional calculus.


问:
How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$
How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$ since $e^0 = 0^0$, and we know the right hand side is undefined?</p>

答:
In this situation, you should view that $0^0$ as

$$\lim_{x \to 0} x^0,$$

which is pretty easily seen to be $1$.

In situations of continuity, it is often better do define $0^0$ as $1$.  One reason I've heard for this is that continuity of $x^0$ is more useful than continuity of the related function $0^x$.


问:
Definition of normed and inner product space
I was reading some Wikipedia pages about Normed Vector spaces and Inner product spaces and, in the definitions, they always talk about vector spaces over either $\Bbb R$ or $\Bbb C$.

Is this because most of the useful normed and inner product spaces are over $\Bbb R$ or $\Bbb C$ or is those spaces only defined for vector spaces over those specific fields?

Edit: After debating this topic in the comments of this post I want to rephrase my question:

Let $V$ be a vector space over a field $\mathbb F$. What condition should $\Bbb F$ verify if we want $V$ to be able to be an inner product space? How about a normed vector space?

答:
I believe it works over any normed field (at least the normed space, for inner product spaces I'm not sure, since you'd need some generalisation for complex conjugation). A normed field $k$ is a field equipped with a norm $||\cdot||: k\to \mathbb{R}_{\ge0}$ such that

$||x||=0\Leftrightarrow x=0$

$||a+b|| \le ||a|| + ||b||$

$||a\cdot b|| = ||a||\cdot||b||$

If your field $k$ has a discrete valuation $\nu$ that you can build a norm by defining $||x||:=\exp(-a\nu(x))$ for any positive $a$...

In any case, I am sure that Bourbaki will provide you with the most general definition.

And if you want to relax the condition that the norm map to $\mathbb{R}_{\ge0}$, I think there is also a way to do that, and just have it map to some kind of totally ordered semiring...


问:
When does a function not have a limit?
What are the cases in which a function does not have a limit? With the exception of piecewise functions, it seems like a function can always be said to have a limit, since I think the value of a function can always be said to lie in the interval $(A-\varepsilon,A+\varepsilon)$.

A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and
only if, for each positive number $\epsilon$ there is another,
$\delta$, such that whenever $0 < |x-a| < \delta $
we have $|f(x) - A| < \epsilon$. That is, when $x$ is near $a$ (within a > distance $\delta$ from it), $f(x)$ is near $A$
(within a distance $\epsilon$ from it).
In symbols we write $\lim_{x \rightarrow a} f(x) = A$.

答:
As a classic example, consider the function
$$
f(x) = \begin{cases}
\sin(1/x) & x \neq 0\\
0 & x = 0.
\end{cases}
$$
The limit $\lim_{x \to 0}f(x)$ does not exist because there is no value of $A$ that works. In particular, define the sequences
$$
x_n = \frac1{2\pi n}, \quad y_n = \frac 1{2 \pi (n + 1/2)}.
$$
Note that we have $f(x_n) = 0$ and $f(y_n) = 1$ for all $n$, but both sequences $x_n,y_n$ approach zero. Now, take $\epsilon = 1/4$, and consider any possible $\delta > 0$ (as small as you'd like). We can always find an $n$ so that $0 < |x_n - 0| < \delta$, so the limit $A$ (if it exists) should satisfy $|A - 0| < \epsilon$. On the other hand, we can also find an $n$ for which $0 < |y_n - 0| < \delta$, so the limit $A$ should also satisfy $|A - 1| < \epsilon$.

However, there is no number $A$ that satisfies this simultaneous requirement of $|A - 0| < 1/4$ and $|A - 1| < 1/4$. So, the function has no limit at $x = 0$.

Title: When does a function not have a limit?


问:
Find all natural triplets $(a, b, c)$ such that $\frac{3}{7} = \frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}$
Find all natural triplets $(a,b,c)$ with property    $a≤b≤c$  that satisfy the equation

$$\dfrac{3}{7} = \dfrac{1}{a}+\dfrac{1}{ab}+\dfrac{1}{abc}$$

答: Hint: $3abc = 7 (bc+c+1)$, so $(3ab-7b-7)c = 7$.  What could $c$ possibly be?


问:
No. of ways 7 digit Phone Number can be obtained
I'm trying to solve a problem based on counting.
How many ways are there to create a 7 digit phone no provided the following two conditions are met:

1.The first digit cannot be 0 or 1.

2.The first three-digits cannot be 911.

I tried to solve it using the multiplication rule: the first digit can be selected in 7 ways (since 0,1 and 9 not allowed) and the second digit in 9 ways (since 1 not allowed) and the third digit also 9 ways (since 1 is not allowed)
and the answer I evaluated came up to $7.9.9.10^4$.

But the solution given online seems to be different.
They have evaluated the first part as $8.10^6$(without 0 or 1 as the first digit) and
then they evaluated the second part as :since 9 1 1 can be chosen in only one and the rest 4 digits can be chosen $10^4$ ways, the no of ways is $8.10^6-10^4$ which does not match my answer.

Where is my logic going wrong?

Note that there is a strange thing of sorts with my answer and the online solution. My solution is $7.9.9 .10^4$ and online solution is $799.10^4$. Is there something here that I'm not seeing? Or is this just coincidence?

答: Your logic is wrong in that the first digit can be $9$. One specific case is $9999999$ which is allowed. What you need to do is have $8\times10^6$ and minus the case where the first three digits are $911$ in that order, which is $10^4$. And yes, the solution is just a coincidence.


答:
I think you misunderstand that not just one value of $f(x)\in (A-\epsilon, A+\epsilon)$.  But ALL $x\in (a-\delta, a+\delta)$ will be that $f(x)\in (A-\epsilon, A+\epsilon)$. It's possible that $x_1 \in (a-\delta, a+\delta)$ and $x_2\in (a-\delta, a+\delta)$ but $f(x_2) \not \in (A-\epsilon, A+\epsilon)$.

Consider $f(x) = 0$ if $x $ is rational and $f(x) = 1$ if $x$ is irrational.

Is it possible that $\lim_{x\to 0} f(x) = A$?

The answer is no, the limit doesn't exist.

Let $\epsilon = 0.1$.  If $\lim_{x\to 0} f(x) = A$ then there would be a $\delta$ so that if $x \in (-\delta, \delta)$ then $|f(x) - A| < 0.1$.

Suppose so.  suppose $x \in (-\delta, \delta)$ is rational. that would mean $|f(x)-A| = |0-A| = |A| < 0.1$ and $-0.1 < A < 0.1$.

But suppose $x_2 \in (-\delta, \delta)$ is irrational.  That wouldd mean $|f(x) -A| = |1-A|=|A-1| < 0.1$ and $-0.1 < A-1 < 0.1$ and $0.9 < A < 1.1$.

But that means $-.1< A< .1$ and $0.9 < A < 1.1$ and that's a contradiction.

So no limit of $\lim_{x\to 0} f(x)$ exists.

Title: When does a function not have a limit?

Body:
What are the cases in which a function does not have a limit? With the exception of piecewise functions, it seems like a function can always be said to have a limit, since I think the value of a function can always be said to lie in the interval $(A-\varepsilon,A+\varepsilon)$.

A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and
only if, for each positive number $\epsilon$ there is another,
$\delta$, such that whenever $0 < |x-a| < \delta $
we have $|f(x) - A| < \epsilon$. That is, when $x$ is near $a$ (within a > distance $\delta$ from it), $f(x)$ is near $A$
(within a distance $\epsilon$ from it).
In symbols we write $\lim_{x \rightarrow a} f(x) = A$.


答: First digit cannot be '0', so 100 numbers are out. First digit cannot be '1', another 100 numbers are out. Among the remaining 800 numbers, only '911' is not allowed, so you can choose between 799 numbers for the first three digits.</p>

Title: No. of ways 7 digit Phone Number can be obtained

Body:
I'm trying to solve a problem based on counting.
How many ways are there to create a 7 digit phone no provided the following two conditions are met:

1.The first digit cannot be 0 or 1.

2.The first three-digits cannot be 911.

I tried to solve it using the multiplication rule: the first digit can be selected in 7 ways (since 0,1 and 9 not allowed) and the second digit in 9 ways (since 1 not allowed) and the third digit also 9 ways (since 1 is not allowed)
and the answer I evaluated came up to $7.9.9.10^4$.

But the solution given online seems to be different.
They have evaluated the first part as $8.10^6$(without 0 or 1 as the first digit) and
then they evaluated the second part as :since 9 1 1 can be chosen in only one and the rest 4 digits can be chosen $10^4$ ways, the no of ways is $8.10^6-10^4$ which does not match my answer.

Where is my logic going wrong?

Note that there is a strange thing of sorts with my answer and the online solution. My solution is $7.9.9 .10^4$ and online solution is $799.10^4$. Is there something here that I'm not seeing? Or is this just coincidence?


问:
If two closed simple curves have the same trace, same starting point and same starting derivative, then they are equal.
This is not a textbook problem, at least that I'm aware of. I was doodling around with some curves and came to this reasonable conjecture: Let $A\subseteq\mathbb{R}^n$ and $\gamma, \bar\gamma:[a, b]\rightarrow\mathbb{R}^n$ be two regular simple closed curves (so, both are of class $C^{\infty}$ with non vanishing derivative, injective in $[a, b)$, and $\gamma^{(k)}(a)=\gamma^{(k)}(b)$, $\bar\gamma^{(k)}(a)=\bar\gamma^{(k)}(b)$, $\forall k\geq 0$) such that $\gamma([a, b])=A=\bar\gamma([a, b])$, $||\gamma '(t)||=||\bar\gamma '(t)||$ $\forall t\in [a, b]$, $\gamma (a)=\bar\gamma(a)$ and $\gamma ' (a)=\bar\gamma ' (a)$, then $\gamma=\bar\gamma$. I put the last hypothesis so that $\gamma$ and $\bar\gamma$ travel $A$ in the same direction. I still don't know how to attack this problem or even if these hypothesis are enough, so I would appreciate any help with this. Thank you.

答:
HINT:

Like Ted Shifrin suggested, the  idea would be that the length of the curve from $\gamma(a)$ to $\gamma(x)$ equals
$$\int_{a}^x \|\gamma'(t)\| dt$$
(this because the function is injective)
so it equals the length from $\gamma(a)= \bar\gamma(a)$ to $\bar\gamma(x)$. Now if they start in the same direction, it follows that $\gamma(x) = \bar\gamma(x)$, for all $x$.


问:
Equivalence relation of atoms
If $(X,\mathcal{S},\mu)$ is a measure space, a set $A\in\mathcal{S}$ is called an atom of $\mu$ iff $0<\mu(A)<\infty$ and for every $C\subset A$ with $C\in\mathcal{S}$, either $\mu(C)=0$ or $\mu(C)=\mu(A)$.

For two atoms $A$ and $B$, define a relation $A\approx B$ iff $\mu(A\cap B)=\mu(A)$.

For $\mu$ finite, show that $\approx$ is an equivalence relation.

My efforts:

$A\approx A$ since $\mu(A\cap A)=\mu(A)$.

Suppose $A\approx B$. Since $A$ is an atom, $\mu(A)>0$. Since $A\approx B$, $\mu(A\cap B)=\mu(A)>0$. Since $B$ is an atom and $A\cap B\subset B$, either $\mu(A\cap B)=0$ or $\mu(A\cap B)=\mu(B)$. Since $\mu(A\cap B)>0$, we must have $\mu(A\cap B)=\mu(B)$. So $B\approx A$.

Suppose $A\approx B$ and $B\approx C$. We need to show $A\approx C$. Since $C$ is an atom and $A\cap C\subset C$, either $\mu(A\cap C)=0$ or $\mu(A\cap C)=\mu(C)$. We only need to show $\mu(A\cap C)>0$. Assume $\mu(A\cap C)=0$. Then $0<\mu(A)=\mu(A\cap C)+\mu(A\setminus C)=\mu(A\setminus C)$. Since $A\setminus C\subset A\setminus(B\cap C)$, we have $\mu(A\setminus C)\leq\mu(A\setminus (B\cap C))=\mu(A)-\mu(A\cap B\cap C)$.

Then I don't know how to proceed.

答:
Use $\mu((A \cap B) \cup (B \cap C)) = \mu(A \cap B) + \mu(B \cap C) - \mu(A \cap B \cap C)$.  $A \cap B \subseteq B$ and $B \cap C \subseteq B$ so $(A \cap B) \cup (B \cap C) \subseteq B$.  By monotonicity,
\begin{equation*}
0 < \mu(A \cap B) \leq \mu((A \cap B) \cup (B \cap C))
\end{equation*}
so $\mu((A \cap B) \cup (B \cup C)) = \mu(A) = \mu(B) = \mu(C)$.  From this, we conclude that $\mu(A \cap B \cap C) = \mu(A) + \mu(A) - \mu(A) = \mu(A)$ and it follows that $\mu(A \cap C) = \mu(A) = \mu(C)$.

Title: Equivalence relation of atoms


答:
The error in your counting is that it excludes any phone-number that has a 9 in the first digit, a 1 in the second digit, and a 1 in the third digit (ex 9012345 is a valid phone number, but isn't counted in your method). To fix this we could use the Principle of Inclusion and Exclusion alongside your original count.

There are $10^6$ numbers with the first digit as 9, $8\cdot10^5$ numbers with a 1 as the second digit, and $8\cdot10^5$ numbers with a 1 as the third digit. There are $10^5$ numbers with a 9 as the first digit and a 1 as the second and by symmetry, there are $10^5$ numbers with a 9 in the first digit and a 1 in the third digit. There are $8\cdot10^4$ numbers with a 1 in both the first and second digit. We can't have a number with a 9 in the first digit, 1 in the second digit, and 1 in the third because that would violate the 911 restriction.

Adding this information to your original count gives
$10^{6}+2\left(8\cdot10^{5}\right)-2\cdot10^{5}-8\cdot10^{4}+7\cdot9\cdot9\cdot10^{4}=799\cdot10^4$, matching the solution.

As mentioned in the solution, using complementary counting is an easier way to solve this problem, since it's easier to count the numbers that we don't want than count the numbers we do want. There are $10^4$ numbers that violate the 911 restriction, so the number of valid numbers can be counted by $8\cdot10^6-10^4=799 \cdot 10^4$ (where the $8\cdot10^6$ comes from the restriction on the first digit).


Body:
Does it exist a recursive formula that approximates $\sum_{k=1}^n k^k\,$?

Defined $\,S_n=\sum_{k=1}^n k^k$, I am looking for a relation $\,S_{n+1}\sim f(n)\cdot S_n\,$, with $\,S_1=1$ and $f(n)\,$ a suitable function of $\,n$.

A trial and error procedure has led me to consider the following function:
$$f(n)=\frac e 2\cdot\Big(2n+1-\log\Big(1+\frac{n-1}{(2n+1)^2}\Big)\Big)$$
Here are some experimental results:

$n\;\;\;\;\;\;\;\;\;\;err_\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;n\;\;\;\;\;\;\;\;\;err_\%$

$1\;\;\;\;-18.45\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,20\;\;\;\;-1.33\%$

$2\;\;\;\;-14.09\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,30\;\;\;\;-0.85\%$

$3\;\;\;\;\;\,-9.70\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,40\;\;\;\;-0.61\%$

$4\;\;\;\;\;\,-7.17\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,50\;\;\;\;-0.47\%$

$5\;\;\;\;\;\,-5.68\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,60\;\;\;\;-0.37\%$

$6\;\;\;\;\;\,-4.70\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,70\;\;\;\;-0.30\%$

$7\;\;\;\;\;\,-4.00\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,80\;\;\;\;-0.25\%$

$8\;\;\;\;\;\,-3.49\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,90\;\;\;\;-0.21\%$

$9\;\;\;\;\;\,-3.08\%\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;100\;\;\;\;-0.18\%$

$10\;\;\,\;-2.76\%$

I shall appreciate every comment and suggestion. Many thanks.

Body:
$$S_n=\sum_{k=0}^{n}k^k$$
$$S_{n+1}^*=(2.71830821508223 n+1.3393100014856814) S_n^*;\;S_1=1$$
The general approximation is
$$S_n^*=1.1286516133652116\cdot 0.367875870164984^{1- n} \;\Gamma (n+0.4926998322172109)$$

$$
\begin{array}{r|r|r|r}
n & S_n & S_n^*& error ( \%)\\
\hline
 1 & 1. & 1. & 0. \\
 2 & 5. & 4.05762 & 18.8476 \\
 3 & 32. & 27.4941 & 14.0809 \\
 4 & 288. & 261.036 & 9.36262 \\
 5 & 3413. & 3187.91 & 6.5951 \\
 6 & 50069. & 47598.2 & 4.9348 \\
 7 & 873612. & 840068. & 3.83968 \\
 8 & 1.76508\times 10^7 & 1.71101\times 10^7 & 3.06369 \\
 9 & 4.05071\times 10^8 & 3.94999\times 10^8 & 2.48655 \\
 10 & 1.04051\times 10^{10} & 1.01926\times 10^{10} & 2.04211 \\
 11 & 2.95717\times 10^{11} & 2.90717\times 10^{11} & 1.69072 \\
 12 & 9.21182\times 10^{12} & 9.0822\times 10^{12} & 1.40705 \\
 13 & 3.12087\times 10^{14} & 3.08423\times 10^{14} & 1.17414 \\
 14 & 1.14241\times 10^{16} & 1.13121\times 10^{16} & 0.980217 \\
 15 & 4.49318\times 10^{17} & 4.45648\times 10^{17} & 0.816844 \\
 16 & 1.88961\times 10^{19} & 1.8768\times 10^{19} & 0.677825 \\
 17 & 8.46136\times 10^{20} & 8.41411\times 10^{20} & 0.558506 \\
 18 & 4.01925\times 10^{22} & 4.00095\times 10^{22} & 0.455327 \\
 19 & 2.01861\times 10^{24} & 2.01123\times 10^{24} & 0.365523 \\
 20 & 1.06876\times 10^{26} & 1.0657\times 10^{26} & 0.286909 \\
\end{array}
$$

The formula is a simple linear regression on the ratios of $20$ sums $s_{k+1}/s_k$
They lie on a almost perfect straight line. I used $1000$ ratios to get the formulae above.

The picture below shows the first $20$

$$
...
$$


问:
On the Pell-like $px^2-qy^2 = 1$ for prime $p,q$
Given any prime of form $p_n = u^2+nv^2$ for non-zero integers $u,v$. Consider,

\begin{aligned}
&p_2x^2-2y^2 = 1\\
&p_3x^2-3y^2 = 1\\
&p_7x^2-7y^2 = 1\\
&p_{11}x^2-11y^2 = 1\\
\end{aligned}

Question 1: Are these four families of Pell-like eqns always solvable in non-zero integers $x,y$?

Ex.

Let $n=11$, and since $53=u^2+11v^2$, then $53x^2-11y^2=1$ is solvable. (Other $p_{11}$ will do.)

Question 2: Is there another $n$?

Let $n=13$, and since $157=u^2+13v^2$, but $157x^2-13y^2=1$ is NOT solvable in the integers.

Have these questions been answered in the literature already?

Body:
There is no solution for $$  401 x^2 - 11 y^2 = 1. $$

j

agy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2
401 0 -11

  0  form            401           0         -11  delta     -6
  1  form            -11         132           5

           0          -1
           1          -6

To Return
          -6           1
          -1           0

0  form   -11 132 5   delta  26
1  form   5 128 -63   delta  -2
2  form   -63 124 9   delta  14
3  form   9 128 -35   delta  -3
4  form   -35 82 78   delta  1
5  form   78 74 -39   delta  -2
6  form   -39 82 70   delta  1
7  form   70 58 -51   delta  -1
8  form   -51 44 77   delta  1
9  form   77 110 -18   delta  -6
10  form   -18 106 89   delta  1
11  form   89 72 -35   delta  -2
12  form   -35 68 93   delta  1
13  form   93 118 -10   delta  -12
14  form   -10 122 69   delta  1
15  form   69 16 -63   delta  -1
16  form   -63 110 22   delta  5
17  form   22 110 -63   delta  -1
18  form   -63 16 69   delta  1
19  form   69 122 -10   delta  -12
20  form   -10 118 93   delta  1
21  form   93 68 -35   delta  -2
22  form   -35 72 89   delta  1
23  form   89 106 -18   delta  -6
24  form   -18 110 77   delta  1
25  form   77 44 -51   delta  -1
26  form   -51 58 70   delta  1
27  form   70 82 -39   delta  -2
28  form   -39 74 78   delta  1
29  form   78 82 -35   delta  -3
30  form   -35 128 9   delta  14
31  form   9 124 -63   delta  -2
32  form   -63 128 5   delta  26
33  form   5 132 -11   delta  -12
34  form   -11 132 5
minimum was   5rep 0 1 disc   17644 dSqrt 132.83071934  M_Ratio  145.8182
Automorph, written on right of Gram matrix:
930589980  -691078313
1056608089  1136194800
 Trace:  2066784780   gcd(a21, a22 - a11, a12) : 1
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$


答:
I'd do it in a different way (each step is justified by the previous one):

$\mu(B\cup C)=\mu(B)+\mu(C)-\mu(B\cap C)=\mu(B\cap C)+\mu(C)-\mu(B\cap C)=\mu(C)$

$\mu((A\cap B)\cup C)=\mu((A\cup C)\cap(B\cup C))\le\mu(B\cup C)=\mu(C)$

$\mu(A\cap B\cap C)=\mu(A\cap B)+\mu(C)-\mu((A\cap B)\cup C)\ge\mu(A\cap B)+\mu(C)-\mu(C)=\mu(A\cap B)=\mu(A)$

$\mu(A)\ge\mu(A\cap C)\ge\mu(A\cap B\cap C)\ge\mu(A)\Rightarrow\mu(A\cap C)=\mu(A)$

Title: Equivalence relation of atoms

Body:
If $(X,\mathcal{S},\mu)$ is a measure space, a set $A\in\mathcal{S}$ is called an atom of $\mu$ iff $0<\mu(A)<\infty$ and for every $C\subset A$ with $C\in\mathcal{S}$, either $\mu(C)=0$ or $\mu(C)=\mu(A)$.

For two atoms $A$ and $B$, define a relation $A\approx B$ iff $\mu(A\cap B)=\mu(A)$.

For $\mu$ finite, show that $\approx$ is an equivalence relation.

My efforts:

$A\approx A$ since $\mu(A\cap A)=\mu(A)$.

Suppose $A\approx B$. Since $A$ is an atom, $\mu(A)>0$. Since $A\approx B$, $\mu(A\cap B)=\mu(A)>0$. Since $B$ is an atom and $A\cap B\subset B$, either $\mu(A\cap B)=0$ or $\mu(A\cap B)=\mu(B)$. Since $\mu(A\cap B)>0$, we must have $\mu(A\cap B)=\mu(B)$. So $B\approx A$.

Suppose $A\approx B$ and $B\approx C$. We need to show $A\approx C$. Since $C$ is an atom and $A\cap C\subset C$, either $\mu(A\cap C)=0$ or $\mu(A\cap C)=\mu(C)$. We only need to show $\mu(A\cap C)>0$. Assume $\mu(A\cap C)=0$. Then $0<\mu(A)=\mu(A\cap C)+\mu(A\setminus C)=\mu(A\setminus C)$. Since $A\setminus C\subset A\setminus(B\cap C)$, we have $\mu(A\setminus C)\leq\mu(A\setminus (B\cap C))=\mu(A)-\mu(A\cap B\cap C)$.

Then I don't know how to proceed.


问:
If a normalized sequence of finite measures converges weakly, does the same hold true for the nonnormalized one?
Let $E$ be a metric space, $(\mu_n)_{n\in\mathbb N}$ be a sequence of finite nonnegative measures on $\mathcal B(E)$ and $\mu$ be a probability measure on $\mathcal B(E)$ s.t. $$\frac{\mu_n}{\mu_n(E)}\xrightarrow{n\to\infty}\mu\tag1$$ with respect to the to the topology of weak convergence of measures.

Assuming that $c:=\sup_{n\in\mathbb N}\mu_n(E)<\infty$, are we able to deduce that the nonnormalized sequence $(\mu_n)_{n\in\mathbb N}$ is convergent as well?

If not, does the other direction of this assertion hold, i.e. can we infer that the normalized sequence (i.e. the left-hand side of $(1)$) is convergent from knowing that the nonnormalized sequence is convergent? It's easy to see that $(\mu_n(E))_{n\in\mathbb N}$ is convergent in that case (since the function constantly equal to $1$ is bounded and continuous). So, this should be possible to obtain.

Remark: Without assuming $c<\infty$, the first implication is clearly wrong. We simply can consider any finite nonnegative measure $\nu$ on $\mathcal B(E)$ and set $\nu_n:=n\nu$ for $n\in\mathbb N$.

答:
By the same argument as presented by Botnakov N. we can even show more:

Let $\mathcal M(E)$ denote the space of finite signed measures on $\mathcal B(E)$ equipped with the total variation norm $\left\|\;\cdot\;\right\|$, $(\mu_t)_{t\in I}$ be a net in $\mathcal M(E)$ with $$c:=\lim_{t\in I}\left\|\mu_t\right\|\ne0\tag2$$ and $\mu\in\mathcal M(E)$ with $$\left(\frac{\mu_t}{\left\|\mu_t\right\|}\right)_{t\in I}\to\mu\tag3$$ with respect to the topology of weak convergence of measures. Then, $$\mu_tf=\left\|\mu_t\right\|\frac{\mu_t}{\left\|\mu_t\right\|}f\xrightarrow{t\in I}c\mu f\;\;\;\text{for all }f\in C_b(E)\tag4;$$ i.e. $$(\mu_t)_{t\in I}\to c\mu\tag5.$$


问:
Find the direction in which the directional derivative of f(x,y), at the point (x,y)=(0,4), has a value of 1.
f(x,y) = 9e^(-0.5x*y)

Find the direction in which the directional derivative of f(x,y), at the point (x,y)=(0,4), has a value of 1.  Please input your answer as a column vector.

When trying to solve i got:
fx ---> (-18)
fy ---> (0)
So does this mean 1 = (-18,0)*(x,y)
--->   1=-18x+0y
--->   x=1/-18
---> (-1/18)
(  0  )
As a column vector shown above

Or
does this mean 1 = (-18,0)*(cos(theta),Sin(theta))
---> 1 = -18 Cos(theta)
---> Theta = 93.2 degrees
Do i give my answer in degrees or radians and how would i show my answer as a column vector

答:
Note that $f(x,y)=9e^{-\frac{1}{2}xy}$ so clearly
\begin{equation*} \nabla f(x,y)=\langle f_{x}(x,y),f_{y}(x,y)\rangle = \left\langle 9e^{-\frac{1}{2}xy}\left(-\frac{1}{2}y\right),9e^{-\frac{1}{2}xy}\left(-\frac{1}{2}x\right)\right\rangle = -\frac{9}{2}\langle ye^{-\frac{1}{2}xy}, xe^{-\frac{1}{2}xy}\rangle\end{equation*}
and it follows that
\begin{equation*} \nabla f(0,4)=-\frac{9}{2}\langle 4e^{0},0\rangle =\langle -18,0\rangle \end{equation*}
and finally, you need to find the direction (i.e. unit vector $\vec{u}=\langle u_{1},u_{2}\rangle$ such that $\nabla f(0,4)\cdot\vec{u}=1$. Then,
\begin{equation*}1=\nabla f(0,4)\cdot\vec{u}=\langle -18,0\rangle\cdot\langle u_{1},u_{2}\rangle = -18u_{1} \end{equation*}
and this implies that $u_{1}=-\frac{1}{18}$ so ultimately, $\vec{u}=\left\langle-\frac{1}{18},u_{2}\right\rangle$ will work for the choice of $u_{2}$ that makes $\vec{u}$ a unit vector. As column vector per your question, this is $\vec{u}=\begin{pmatrix} -\frac{1}{18} \\ u_{2} \end{pmatrix}$.


问:
Is there a constant uniform relationship between arc lengths of a sphere and their corresponding 'descents' of the radius?
Regarding a bowling ball, a billiard ball and a ping-pong ball:

These 3 balls, to me, look and feel to have a uniform curvature.
Their curvature characteristics are always the same everywhere on each ball.

The same uniformity of curvature holds true for any represented circle that I have ever seen.

5 questions:

(1)  Is there a relationship between a measured “arc length” of a sphere or a circle and the corresponding “descent” of that sphere or circle’s radius?

(2)  If so, is the relationship constant between two or more “arc lengths” (having varying lengths) and the corresponding radius “descent” of each “arc length” of that specific sphere or circle?

(3)  If so, is this relationship constant regardless of the size of the sphere or circle?

(4) Would it be correct to describe this relationship in the following words:  “the ‘descent’ of the radius divided by the length of an ‘arc’?

(5) Might this relationship be expressed as a constant real number, such as,  0.63661977236 (2/pi) or, perhaps,  0.31830988618 (1/pi)?

In case my 5 questions (above) are badly or incorrectly worded, let me provide 6 real world examples (below) that may more accurately describe the relationship I am trying to discover.

I have in my hands a ten pin bowling ball that has the standard diameter of  8.5".
On one side of the bowling ball, I mark 2 points, A and B.
Using a piece of string, I measure the surface distance between A and B as being 2.63".
I place the bowling ball on to the hard surface of a flat table.
I position the ball so that point A is at the very top of the top of the ball, at “12 o’clock”.
I rotate the ball so that point B is as close to the table as it can be (keeping A at 12 o’clock).

(1)  What is the straight line distance between point B and the table?

On the other side of the bowling ball, I mark 2 points, C and D.
Using a piece of string, I measure the surface distance between C and D as being 3.94".
I position the ball so that point C is at the very top of the top of the ball, at “12 o’clock”.
I rotate the ball so that point D is as close to the table as it can be (keeping C at 12 o’clock).

(2)  What is the straight line distance between point D and the table?

I have in my hands a billiard ball that has the standard diameter of 2.25".
On one side of the billiard ball, I mark 2 points, E and F.
Using a piece of string, I measure the surface distance between E and F as being 1.09" .
I place the billiard ball on to the hard surface of a flat table.
I position the ball so that point E is at the very top of the top of the ball, at “12 o’clock”.
I rotate the ball so that point F is as close to the table as it can be (keeping E at 12 o’clock).

(3)  What is the straight line distance between point F and the table?

On the other side of the billiard ball, I mark 2 points, G and H.
Using a piece of string, I measure the surface distance between G and H as being 0.55".
I position the ball so that point G is at the very top of the top of the ball, at “12 o’clock”.
I rotate the ball so that point H is as close to the table as it can be (keeping G at 12 o’clock).

(4)  What is the straight line distance between point H and the table?

I have in my hands a ping-pong ball that has the standard diameter of 1.57".
On one side of the ping-pong ball, I mark 2 points, J and K.
Using a piece of string, I measure the surface distance between J and K as being 0.39".
I place the ping-pong ball on to the hard surface of a flat table.
I position the ball so that point J is at the very top of the top of the ball, at “12 o’clock”.
I rotate the ball so that point K is as close to the table as it can be (keeping J at 12 o’clock).

(5)  What is the straight line distance between point K and the table?

On the other side of the ping-pong ball, I mark 2 points, L and M.
Using a piece of string, I measure the surface distance between L and M as being 0.62".
I position the ball so that point L is at the very top of the top of the ball, at “12 o’clock”.
I rotate the ball so that point M is as close to the table as it can be (keeping L at 12 o’clock).

(6)  What is the straight line distance between point M and the table?

I am superficially asking for 6 real number answers expressed in inches.

But what I am really asking is:  Does there exist a real number constant that expresses a uniform relationship between the length of an arc and its 'descent' of the radius and might that constant hold true for all sizes of spheres and circles?

Thank you kindly.

答:
There certainly is a relationship between the arc distance from the "top" of a spherical object and the straight-line distance to the table on which the sphere is placed.
Whether it is "uniform" depends on how you define the relationship.

Take a sphere of radius $r$. Mark two points $P$ and $Q$ on the surface of the sphere. Measure the distance between the points along the shortest path that follows the surface of the sphere (the path the string will take when pulled tight); suppose this distance is $s$.

Put the sphere on top of a flat horizontal plane with $P$ at the very "top" of the sphere; this means that the sphere contacts the plane at a point $P'$ exactly opposite from $P$, that is, the line segment $PP'$ passes through the sphere's center and is a diameter of the sphere.

Note that if you keep the sphere in contact with the plane with $P$ positioned at the "top", you must keep $P'$ in contact with the horizontal plane, and the only way to rotate the sphere under these constraints is to rotate it around the axis $PP'.$
As you do so, the point $Q$ traces a horizontal circle over the horizontal plane, that is, it is always at the same distance from the plane, that is, it is always as close as it can be to the plane.

Now if $O$ is the point at the exact center of the sphere, the angle
$\angle POQ$ is $s/r$ radians.
Let $\theta = s/r.$
If we then take a cross-section of this entire construction in the vertical plane through $P,$ $Q$, $P'$, and $O$, it starts to look like this:

To complete the figure, we drop a perpendicular line from $Q$ to the horizontal plane below, intersecting the plane at $Q'$,
and we put $N$ on the diameter $PP'$ so that $NQ$ is perpendicular to $PP'.$
Let the distance $NO$ be $v$ and let the distance $QQ'$ be $h$;
the distance $h$ is the straight-line distance between $Q$ and the horizontal plane (aka table), which is what you asked about.

A little trigonometry tells us that
$$v = r \cos(\theta) = r \cos\left(\frac sr\right).$$
Also, we can observe that $P'NQQ'$ is a rectangle and that
$NP' = NO + OP' = v + r = h = QQ'.$
So the formula for $h$ is
$$ h = r + v = r + r \cos\left(\frac sr\right). \tag1 $$

Take any of your examples, replace $r$ with the ball's radius and replace $s$ with the length of the string, and then $h$ computed by the formula above will be the straight-line distance from the lower end of the string to the table.

There is not going to be a uniform formula that gets you directly from $s$ to $h$. The formula also must involve $r.$
That is, if you are simply measuring the arc length and "descent" in inches,
every spherical object with a different radius is going to have a different "descent" for a given arc length.

Even if you consider just one radius, there is no constant you can multiply by the string length to find the "descent".
The amount by which the function value $\cos(s/r)$ changes is not proportional to the amount by which $s$ changes.

You can simplify things a little bit if you measure everything relative to the radius of the ball rather than just using the number of inches.
Measured relative to the radius of the ball, the length of the string is
$\theta = s/r$ (that's what we mean when we say the angle is $\theta$ radians).
Let $\delta$ be the distance from $Q$ to the horizontal plane below,
measured relative to the radius of the ball;
then $\delta = h/r.$
Dividing both sides of Equation $(1)$ above by $r$ we find that

$$ \frac hr = 1 + \cos\left(\frac sr\right), $$

that is,

$$ \delta = 1 + \cos(\theta). $$

This is about as simple a relationship as you will be able to find between the arc length from $P$ to $Q$ and the height of $Q$ above the plane.


问:
Find $\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x$
Evaluate $$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x$$
My attempt:
Consider
$$I=\int_{0}^{1}x^ne^{x^2}$$
using Taylor's series expansion for $e^{x^2}$ we get
$$I=\int_{0}^{1}x^n\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots\right)$$
So we get
$$I=\int_{0}^{1}x^n+\int_{0}^{1}\frac{x^{n+2}}{1!}+\int_{0}^{1}\frac{x^{n+4}}{2!}+\cdots$$
Integrating we get
$$I=\frac{1}{0!(n+1)}+\frac{1}{1!(n+3)}+\frac{1}{2!(n+5)}+\cdots$$
Now we have,
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=\lim_{n \to \infty}nI$$
Thus we obtain
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=\lim_{n \to \infty}\frac{n}{0!(n+1)}+\frac{n}{1!(n+3)}+\frac{n}{2!(n+5)}+\cdots=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+..$$
So we get
$$\lim _{n \rightarrow \infty} \int_{0}^{1} n x^{n} e^{x^{2}} d x=e$$
is this a valid approach?

答:
Here is yet another approach, based on an observation that the integral is concentrated near $1$.

Fix any $0 < \delta < 1 $ and write
$$
I_n := \int\limits_0^1 n x ^n e^{x^2} dx = \int_0^\delta + \int_{1 - \delta}^1 =: I_1 + I_2.
$$

Clearly
$$
\tag{1} I_1 \leq e^{\delta^2} \int\limits_0^\delta n x^n dx = e^{\delta^2} \frac{n}{n+1} \delta^{n+1} \to 0 , \text{ as } n \to \infty.
$$

For the second part, in view of the monotonicity of $e^{x^2}$ we get
$$
e^{(1-\delta)^2} \frac{n}{n + 1} \left( 1 - (1-\delta)^{n+1}\right) = e^{(1-\delta)^2} \int\limits_{1-\delta}^1 n x^n dx\leq I_2 \leq e \int\limits_0^1 n x^n dx = e \frac{n}{n+1}.
$$

Thanks to $(1)$ and taking $n\to \infty$ in the last inequalities, we get that
$$
e^{(1-\delta)^2} \leq \liminf I_n \leq \limsup I_n \leq e
$$
for any $0 < \delta < 1$. Passing to limit in the last expression as $\delta \to 0$ we prove that the limit of $I_n$ exists as $n\to \infty$ and equals $e$.


问:
Confused with a proof about Galois theory
So I am self-studying Galois theory and I am currently reading about the fundamental theorem of Galois theory. There is a certain lemma/theorem that is used in the proof of the fundamental theorem which I can not seem to understand the proof of. I have been using this website https://jhavaldar.github.io/notes/2017/10/19/galoistheory1.html (1) to read about it. However, the proof of theorem 2 is omitted so I looked around in my bookmarks and found this source: http://www-groups.mcs.st-andrews.ac.uk/~martyn/5836/MT5836lecturenotes.pdf (2). Here, theorem 2  in the source (1) is equivalent to lemma 6.9 (page 56) in the source (2). The lemma goes as follows:

Lemma: Let $K$ be a finite separable extension of a field $F$ and let $H$ be a finite group of $F$-automorphisms of $K$ (that is, $H$ is some subgroup of $Gal(K/F)$). Then $[K:H^*] = |H|$
where $H^* =$ Fix$_K(H)$.

Proof found in source (2): Write $L = H^*$. By the theorem of the primitive element $K = F(\alpha)$ for some $\alpha \in K$. Put $g(X) = \prod_{\varphi \in H} (X - \varphi(\alpha)).$
The proof goes on and shows that $g(X) \in L[X]$.
The definition of $g(X)$ says that it splits in $K$ and, since $K = F(\alpha)$, we certainly build $K$ by adjoining the roots of $K$ to the subfield $L$. Thus $K$ is the splitting field of $g(X)$ over $L$...

The last part that is highlighted in italic is the part that I cant seem to grasp. I know that $g(X)$ splits in $K$ by definition, but I do not understand how we build $K$ by adjoining the roots of $K$ to the subfield $L$. What does it mean to adjoin roots of $K$ to a field? How can this show that $K$ is a splitting field for $g(X)$, i.e it is the minimal field in which $g(X)$ splits in?

答:
When you adjoin an element $\alpha$ to a field $F$, we are talking about the smallest field containing $F$ and $\alpha$. If $\alpha$ is algebraic of degree $n$ over $F$, this is spanned as a vector space over $F$ by $1, \alpha, ..., \alpha^{n-1}$.

In your question, $K = F(\alpha)$. Let $\alpha_1 = \alpha$ and write $\alpha_1,..., \alpha_n$ for the roots of $g$. By definition, $\{\alpha_1,..., \alpha_n\} \subseteq K$ and also $L \subseteq K$. Hence, we have that $L(\alpha_1,..., \alpha_n) \subseteq K$.

On the other hand, $K = F(\alpha)$ and $L \supseteq F$ so $L(\alpha_1,..., \alpha_n) \supseteq L(\alpha) \supseteq F(\alpha) = K$.

So $L(\alpha_1,..., \alpha_n) = K$ and it must be the splitting field of of $g(x)$.


答:
The idea is that $A\approx B$ iff the difference between $A$ and $B$ is negligible (in measure). We do just that: Suppose $A\approx B$ and $B\approx C$.

Notice that
\begin{align*}
A\setminus C
&\subseteq ((A\setminus B)\cup B)\setminus C\\
&=\left[(A\setminus B)\setminus C\right]\cup \left[B\setminus C\right]\\
&\subseteq (A\setminus B)\cup (B\setminus C)
\end{align*}
so $\mu(A\setminus C)\leq\mu(A\setminus B)+\mu(B\setminus C)$.

Now we have the decomposition of $A$ as a disjoint union $A=(A\setminus B)\sqcup (A\cap B)$, so from $A\approx B$ we get
$$\mu(A)=\mu(A\setminus B)+\mu(A\cap B)=\mu(A\setminus B)+\mu(A),$$
so $\mu(A\setminus B)=0$. Similarly, from $B\approx C$ you get $\mu(B\setminus C)=0$.

Then
$$0\leq\mu(A\setminus C)\leq\mu(A\setminus B)+\mu(B\setminus C)=0$$
Therefore $\mu(A\setminus C)=0$. Now we decompose $A=(A\setminus C)\sqcup(A\cap C)$ to obtain
$$\mu(A)=\mu(A\setminus C)+\mu(A\cap C)=0+\mu(A\cap C)=\mu(A\cap C),$$
which means that $A\approx C$.

So in the end, what I'm doing is using the following lemma (which follows by decomposing $A=(A\setminus B)\cup (A\cap B)$) a few times, which rewrites $\approx$ in terms of differences and makes some computations easier:

For atoms $A,B$, the following are equivalent:

$A\approx B$;

$\mu(A\setminus B)=0$.

Title: Equivalence relation of atoms


问:
Is this braced heptagon a rigid graph?
In Mathematica, GraphData[{"UnitDistance", {21, 2}}]

Is this 42-edge graph rigid?  It has chromatic number 4. If it was floppy, that might make it an interesting tool in high-chromatic graph studies.

The current record for heptagon bracing is 59, so this would beat that.

答: My proof was written up in a Mathematica. I've made a copy available on Google-Drive both as a Mathematica notebook  (NewHeptagon-Draft1.nb) and as a static pdf (NewHeptagon-Draft1.pdf). Since the version of Mathematica I was using didn't have GraphData[{"UnitDistance", {21, 2}}] I had to regenerate the graph from just its picture so that's embedded in there as well.

Title: Is this braced heptagon a rigid graph?


问:
How to show $ \int_0^T\int_\Omega \frac{\partial }{\partial t}( \varphi^2 ) \,dx \,dt = \int_0^T \frac{d}{dt} (\int_\Omega \varphi^2 \, dx) \,dt $?
$\Omega\subset \Bbb{R}^n$ is a bounded smooth domain.

$Q_T=\Omega\times [0,T]$.

$\mathring C^\infty(Q_T)=\{u\in C^\infty(Q_T) : u \text{ is zero near } \partial\Omega\times[0,T]\}$.

$\mathring W^{1,1}_2(Q_T)$ is closure of $\mathring C^\infty(Q_T)$ in $W^{1,1}_2(Q_T)$.  $W^{1,1}_2(Q_T)$ is Sobolov space, $u\in W^{1,1}_2(Q_T)$ means $u_t$ and $u_x$ belong to $L^2(Q_T)$.

For any $\varphi\in \mathring W^{1,1}_2(Q_T)$, how to show
$$
\int_0^T\int_\Omega \frac{\partial }{\partial t}( \varphi^2 ) \, dx \, dt = \int_0^T \frac{d}{dt} \left(\int_\Omega \varphi^2 \,dx\right)\, dt
$$

答:
I will outline the proof but will let you to fill in the details for by the level of the proposition I presume you are able to.

Writing $\psi := \varphi^2$ to simplify notation, the first thing to note is that the outer integrals are equal
$$\int_0^T \left( \int_\Omega \frac{\partial \psi}{\partial t}\ dx \right) dt = \int_0^T \frac{d}{dt} \left( \int_\Omega \psi\ dx \right)\ dt$$
if and only if the inner integrals are equal almost everywhere with respect to Lebesgue measure
$$\int_\Omega \frac{\partial \psi}{\partial t}\ dx = \frac{d}{dt} \int_\Omega \psi\ dx\ \text{ a.e.}$$

That is, you need to show that the "integral with respect to $x$ and the derivative with respect to $t$ commutes". This is done through the Dominated Convergece Theorem on the usual Lebesgue measure space applied to the sequence of functions
$$f_n(x, t) := \frac{\psi\left(x, t + \frac{1}{n}\right) - \psi(x, t)}{\frac{1}{n}}$$

For this you should argue that this is indeed a sequence of Lebesgue measurable functions in $x$ and also
$$\lim_{n \to \infty} f_n = \frac{\partial \psi}{\partial t}$$

Next you argue that since $\psi$ vanishes when close enough to the boundary then also $\frac{\partial \psi}{\partial t}$ does, and since $\Omega$ is bounded and connected then $\left| \frac{\partial \psi}{\partial t} \right|$ attains its maximum.

Then by the Mean Value Theorem with respect to $t$ you should argue that fixed $t_0 \in [0, T]$, for all $n \in \mathbb{N}$ and $x \in \Omega$ it holds that
$$|f_n(x, t_0)| \leq \max_{\omega \in \Omega} \left| \frac{\partial \psi}{\partial t}(\omega, t_0) \right|$$
which is constant and therefore Lebesgue-integrable with respect to $x$ on a bounded set. Thus concluding the proof.


问:
Inverse Fourier transform of $\frac{\frac{1}{1+2i\pi f}+\frac{1}{3+2i\pi f}}{\frac{2}{1+2i\pi f}-\frac{2}{4+2i\pi f}}$
I wanted to calculated the inverse fourier transform of the folowing function :
$\frac{\frac{1}{1+2i\pi f}+\frac{1}{3+2i\pi f}}{\frac{2}{1+2i\pi f}-\frac{2}{4+2i\pi f}}$
($f$ is the variable)

I tried the exponential form, Unifying the denominators but in vain, the premitive is still impossible to be found !

can I get some help ?  much appreciated.

答:
After simplifying, we have $$H(f) =\frac{\frac{1}{1+2i\pi f}+\frac{1}{3+2i\pi f}}{\frac{2}{1+2i\pi f}-\frac{2}{4+2i\pi f}} = 1 + \frac{2 i f \pi}{3} + \frac{i}{-9 i + 6 f π}$$
Now use these facts $$\mathcal{F}\{g'(x)\} = 2\pi i f G(f)$$ $$\mathcal{F}\{e^{-ax}u(x)\} = \frac{1}{a + 2\pi i f}$$ $$\mathcal{F}\{\delta(x)\} = 1$$
You can check your result(Be careful about the scaling factors which are result of how the Fourier transform is defined).


问:
Elementary proof that $\mathbb{Q}$ is not a finitely presented $\mathbb{Q}[x_1,x_2,\dots]$-module
I am trying to prove in an elementary way that $\mathbb{Q}$ is not a finitely presented $\mathbb{Q}[x_1,x_2,\dots]$-module.

According to this post, the general case for any non noetherian ring can be proven by showing that $\mathbb{Q}$ being finitely presented implies that $(x_1,x_2,\dots)$ would be a finitely generated ideal in $\mathbb{Q}[x_1,x_2,\dots]$, which is clearly false. However the proof uses Schanuel's lemma which I am not familiar with.

Therefore I am wondering whether there is a more elementary proof to show the result for the $\mathbb{Q}[x_1,x_2,\dots]$-module $\mathbb{Q}$?

答:
Assume that $\mathbb{Q}$ is finitely presented. Then there exists $m \in \mathbb{N}$ and a finitely generated submodule $K \subset \mathbb{Q}[x_1,x_2,\dots]^m$, s.t. $\mathbb{Q} \cong \mathbb{Q}[x_1,x_2,\dots]^m /K$. Let $\varphi: \mathbb{Q} \to \mathbb{Q}[x_1,x_2,\dots]^m /K$ be such an isomorphism.

$\varphi$ is bijective and $\mathbb{Q}[x_1,x_2,\dots]$-linear, thus it is also $\mathbb{Q}$-linear, hence a $\mathbb{Q}$-linear vector space isomorphism. Thus $\dim_\mathbb{Q}(\mathbb{Q})=\dim_\mathbb{Q}(\mathbb{Q}[x_1,x_2,\dots]^m /K)$.

Since $K$ is finitely generated there exist $i_1,\dots,i_n \in \mathbb{N}$ and $f_1,\dots,f_s \in\mathbb{Q}[x_{i_1},\dots,x_{i_n}]^m$ with $K= \langle f_1,\dots,f_s\rangle$. Choose $x_j \notin \{x_{i_1},\dots,x_{i_n}\}$. Then $[(x_j,0,\dots,0)]$ and $[(x_j^2,0,\dots,0)]$ are $\mathbb{Q}$-linearly independent elements in $\mathbb{Q}[x_1,x_2,\dots]^m /K$ and so $\dim_\mathbb{Q}(\mathbb{Q}[x_1,x_2,\dots]^m /K) \geq 2$, which contradicts $\dim_\mathbb{Q}(\mathbb{Q})=1$.


问:
Generalization of the concept of Cartesian product in Halmos' book.
On page 36 of his Set Theory book, Halmos states the following (with $X$ and $Y$ being two sets whose cartesian product we'll consider):

"Consider any particular unordered pair {a, b} with $a\neq b$, and consider the set Z of all families z, indexed by {a, b}, such that $z_a \in X$ and $z_b \in Y$. If the function $f$ from $Z$ to $X \times Y$ is defined by $f(z) = (z_a, z_b)$, then $f$ is a one-to-one correspondence between $X \times Y$ and $Z$."

My question is: where is this one-to-one correspondence? The way I'm interpreting this is the following:

We have a set of families $Z$, which is the range of some family {$ z_i $}, and where each term of this family ($Z$) is also a family. Then, the domain of such family {$ z_i $} is the index $I=\, ${$a, b$}; such that the terms of the family will be $z_a$ and $z_b$. Then, if $z_a \in X$ and $z_b \in Y$, if we define a function $f: Z\longrightarrow X\times Y$, where $f(z):= (z_a, z_b)$, then, whatever $z$ we choose from $Z$, will its image be the fixed ordered pair $(z_a, z_b)$?

Sorry if I'm overcomplicating this, I just started studying families and this has me a bit confused.

Thanks in advance.

答:
A family of elements indexed by the two element set $\{a,b\}$ is just a function with domain $\{a,b\}$, and for such a function $z$, the index notation simply means $z_a=z(a),\ z_b=z(b)$. So the set $Z$ is
$$Z=\{z:\{a,b\}\to X\cup Y: z(a)\in X,\,z(b)\in Y\}\,.$$
The correspondence to $X\times Y$ is given by $z\mapsto (z(a),\,z(b))$ and $(x,y)\mapsto (a\mapsto x,\ b\mapsto y)$.


问:
Double integral over a typical region.

Question: Let $R$ be the region in $xy$-plane bounded by the curves $y=x^2,y=4x^2,xy=1,xy=5$.
Then find the value of line integral $\iint_R\frac{y^2}{x}dydx$.

I found it tough and time taking if I evaluate this integral from general method, is there any theorem which make it easy? Since the question is asked in competitive exam so I hope there must be an easy and fast way to solve it.

Any help will be appreciable.

Thanks in advance.

答:
Try a change of variables by letting $u=x^2/y$ and $v=xy$. Then the region $R$  is transformed into the rectangle $[1/4,1]\times [1,5]$. Since the Jacobian determinant is
$$\det\left(\begin{bmatrix}
\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}
\end{bmatrix}\right)
=\det\left(\begin{bmatrix}
2x/y& -x^2/y^2\\
y& x
\end{bmatrix}\right)=\frac{3x^2}{y},$$
it follows that
$$\iint_R\frac{y^2}{x}dydx=\frac{1}{3}\int_{u=1/4}^1\int_{v=1}^5\frac{y^3}{x^3}dvdu=
\frac{1}{3}\int_{1/4}^1\frac{du}{u^2}\cdot\int_{1}^5vdv.$$
Can you take it from here?

Title: Double integral over a typical region.


问:
Finding Cartesian coordinates of remaining vertices of triangle, given a vertex and an angle
I have an isosceles triangle $ABC$, where the height $h$ and the angle at vertex $A$ are known. The Cartesian coordinates of vertex $A$ are also known to be $\left(x,y\right)$. is it possible to find the coordinates of the remaining vertices?

Triangle ABC

答:
It is not possible to find the coordinates of the remaining two vertices.

Rotate the triangle about A. The height remains the same. Angle A remains the same. The coordinates of A remain the same but the coordinates of B and C change.

Therefore, there is not enough information to find the coordinates of the remaining vertices.


问:
Functor From Category of Covering Spaces to Category of Sets Equipped With An Action By The Fundamental Groupoid
I have some problems with the understanding of the connection between covering spaces and the fundamental groupoid.

Let $X$ be a topological space and let $\Pi_1(X)$ denote the fundamental groupoid of $X$. We can then construct the category $\Pi_1(X)-\mathbf{Sets}$, which has functors $F:\Pi_1(X)\to\mathbf{Sets}$ as its objects and natural transformations $\eta:F\Longrightarrow G$ as its morphisms ($F,G\in\Pi_1(X)-\mathbf{Sets}$). Let furthermore $\mathbf{Cov}(X)$ be the category of covering spaces over $X$. I would like to construct a functor
$$\Phi:\mathbf{Cov}(X)\to\Pi_1(X)-\mathbf{Sets}.$$
I think I have managed to construct the functors correctly, but not the morphisms.

Construction of Functors

Let $p:Y\to X$ be a covering space, we define a functor
$$F:\Pi_1(X)\to\mathbf{Sets}$$
as follows.

Objects of the fundamental groupoid are points of $X$, so we simply define
$$F(x)=p^{-1}(x),$$
where $x\in X$.

A morphism in $\Pi_1(X)$, between two points $x_0$ and $x_1$, is a path $\alpha:x_0\to x_1$ modulo homotopy equivalence. Given such a path, we construct a set-theoretical map $\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1)$ (from a point $y_0\in p^{-1}(x_0)$ to a point $y_1\in p^{-1}(x_1)$) as follows:

We have a covering space $p:Y\to X$, which makes it possible to apply the path-lifting property. That is, we lift $\alpha:x_0\to x_1$ to a path $\widetilde{\alpha}:y_0\to y_1$, such that $\beta(0)=y_0$ and $p\widetilde{\alpha}=\alpha$. Letting $\widetilde{\alpha}(1)=y_1$, one can check that $p\widetilde{\alpha}(1)=p(y_1)=\alpha(1)=x_1$, which gives a well-defined map
$$\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1).$$
One can then check that this satisfies the functorial properties.

Construction of Morphisms

So, I think I know how to do this step without any reference to functors (it feels like it should be an easy translation to functors, but somehow I cannot figure it out). Let us begin to do this without any functors.

A Construction Without Functors

Let $X$ be a topological space. Without the category-theoretical language, I think this is how the argument goes.

Let $(Y_0,p_0)$ and $(Y_1,p_1)$ be covering spaces of a topological space $X$. A morphism between the two covering spaces is then a map $\varphi:Y_0\to Y_1$ such that $p_1\circ \varphi=p_0$.

We want to use the covering space morphism to construct a $\pi_1(X,x)-\text{Set}$ morphism; That is, maps between sets which preserves the action by the paths. What one can do is to simply combine the covering space morphism and the path-lifting diagram to show this.

Let $\alpha:I\to X$ be a path, by the path lifting property, we can lift $\alpha$ to a path $\widetilde{\alpha}:I\to Y$ such that ${p_0}_*(\widetilde{\alpha})=\alpha$ and $\widetilde{\alpha}(0)=y$, with $y\in p_0^{-1}(x)$. Thus $$\alpha_*(y)=\widetilde{\alpha}(1).$$
We get another path through $\varphi_*(\widetilde{\alpha})$ in $Y$, with initial point $\varphi_*(\widetilde{\alpha})(0)=\varphi(y)$ and terminal point $\varphi_*(\widetilde{\alpha})(1)=\varphi(\alpha_*)(y)$.

The following extended commutative diagram shows that $p_*[\varphi_*(\widetilde{\alpha})]=\alpha$ (with an extra map $p_0:Y_0\to X$, but I didn't knew how to include it, without making a total mess of the diagram).
$\require{AMScd}$
\begin{CD}
Y_0 @>\varphi>> Y_1\\
@AA\widetilde{\alpha}A @VVp_1V\\
I @>\alpha>> X
\end{CD}
Hence $\varphi_*(\widetilde{\alpha})$ is a lifting of $\alpha$. So we get that the terminal point of $\alpha_*(\varphi y)$ is the terminal point of $\varphi_*(\widetilde{\alpha})$, also. So
$$\alpha_*(\varphi(y))=\varphi(\alpha_*(y)).$$
Thus, we have a morphism between path-actions.

A Construction With Functors

Given a covering space homomorphism $\varphi:(Y_0,p_0)\to (Y_1,p_1)$, we want to map it to a morphism in $\Pi_1(X)-\mathbf{Sets}$, by applying $\Phi$. In a category of functors a morphism is a natural transformation.

A natural transformation is constructed as follows. Given two functors $\Gamma,\Delta:\mathcal{A}\to\mathcal{B}$ between two categories $\mathcal{A},\mathcal{B}$, we have morphism $\mu_X:\Gamma(X)\to\Delta(X)$, and a commutative diagram
$\require{AMScd}$
\begin{CD}
\Gamma(X) @>\mu_X>> \Delta(X)\\
@VV\Gamma(f)V @VV\Delta(f)V\\
\Gamma(Y) @>\mu_Y>> \Delta(Y),
\end{CD}
where $f:X\to Y$ is a morphism in $\mathcal{A}$.

In our case, we have two functor $F,G:\Pi_1(X)\to\mathbf{Sets}$, morphisms $\eta_x:F(x)\to G(x)$ and a commutative diagram
$\require{AMScd}$
\begin{CD}
F(x_0) @>\eta_{x_0}>> G(x_0)\\
@VVF(\alpha)V @VVG(\alpha)V\\
F(x_1) @>\eta_{x_1}>> G(x_1),
\end{CD}
with $\alpha:x_0\to x_1$ a path.

Question 1. What does the natural transformation diagram look like? Is it correct to rewrite it as
$\require{AMScd}$
\begin{CD}
p_0^{-1}(x_0) @>\eta_{x_0}>> p_1^{-1}(x_0)\\
@VVF(\alpha)V @VVG(\alpha)V\\
p_0^{-1}(x_1) @>\eta_{x_1}>> p_1^{-1}(x_1),
\end{CD}
or do we get something like
$\require{AMScd}$
\begin{CD}
p_0^{-1}(x_0) @>\eta_{x_0}>> p_0^{-1}(x_0)\\
@VVF(\alpha)V @VVG(\alpha)V\\
p_1^{-1}(x_1) @>\eta_{x_1}>> p_1^{-1}(x_1)?
\end{CD}
Question 2. How do we define $\eta_x:F(x)\to G(x)$? If $F(x)=p_0^{-1}(x)$ and $G(x)=p_1^{-1}(x)$, then we want to construct a map $\eta_x:p_0^{-1}(x)\to p_1^{-1}(x)$. I would like to construct this map through $\varphi$, I guess. But I'm not really sure how to do it.

Question 3. If $F(x_0)=G(x_0)=p_0^{-1}(x_0)$. Don't we have the following equality then $F(\alpha)=G(\alpha)$? Which I think maybe talks against on of my suggestions in Question 1.

I think all I should do, is to try to mimic what we did in the previous subsection. Or well, if not, more importantly just how to construct the natural transformation. But I don't really know what I am doing at all, there are just so much abstract things happening at the same time right now - which makes me confused.

I would be really happy if someone could help me out with the expression of the natural transformation.

Best wishes,

Joel

答:
For question 1, your first diagram is correct.

For question 2, the map $\eta_x: p_0^{-1}(x) \to p_1^{-1}(x)$ is just the restriction of the covering space homomorphism $\phi: Y_0 \to Y_1$ to $p_0^{-1}(x)$.  The image lands in $p_1^{-1}(x)$ since $\phi$ preserves the projections.

To show that the naturality square commutes for $\eta$ constructed this way, use the uniqueness of path-lifting for covering spaces.


问:
Definition : The Distribution of laplacian of a function?
Let $u\in H_0^1(\Omega)$(but i think we can be way more general than than).

How is $\langle\Delta u,\phi\rangle$ (distribution) defined for $\phi \in D(\Omega)$ a test function ?

答:
$$
\langle \Delta u, \phi \rangle := \langle u, \Delta\phi \rangle.
$$


问:
Solving a 2nd order ODE
Consider
$x''-2x'+x= te^t$

Determine the solution with initial values $x(1) = e,$  $x'(1) = 0.$

I know this looks like and probably is a very easy question, but i'm not getting the right answer when i try and solve putting into quadratic form. Could someone please demonstrate or show me a different method?

Many thanks :)

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\on{x}''\pars{t} -2\on{x}'\pars{t} + \on{x}\pars{t} = t\expo{t}}\,,\qquad
\begin{array}{c}
\mbox{Initial Conditions}
\\
\left\{\begin{array}{rcl}
\ds{\on{x}\pars{1}} & \ds{=} & \ds{\expo{}}
\\[1mm]
\ds{\on{x}'\pars{1}} & \ds{=} & \ds{0}
\end{array}\right.\end{array}}$

$\ds{\pars{\totald{}{t} - 1}\
\overbrace{\pars{\totald{}{t} - 1}
\on{x}\pars{t}}^{\ds{\on{y}\pars{t}}}\ =\ t\expo{t}}$

\begin{align}
&\totald{\bracks{\expo{-t}\on{y}\pars{t}}}{t} = t
\\[2mm]
&\expo{-t}\on{y}\pars{t} = {1 \over 2}\,t^{2} + a \implies
\on{y}\pars{t} =
{1 \over 2}\,t^{2}\expo{t} + a\expo{t}\,\quad a = \mbox{constant}
\end{align}

$\ds{\pars{\totald{}{t} - 1}\on{x}\pars{t} =
{1 \over 2}\,t^{2}\expo{t} + a\expo{t}}$

\begin{align}
&\totald{\bracks{\expo{-t}\on{x}\pars{t}}}{t} =
{1 \over 2}\,t^{2} + a
\\[2mm] &\
\expo{-t}\on{x}\pars{t} = {1 \over 6}\,t^{3} + at + b\,,\quad b = \mbox{constant}
\\[2mm] &\
\on{x}\pars{t} = \pars{{1 \over 6}\,t^{3} + at + b}\expo{t}
\\[2mm] &\
\left\{\begin{array}{rcrcl}
\ds{\expo{}a} & \ds{+} & \ds{\expo{}b} & \ds{=} &
\ds{{5 \over 6}\,\expo{}}
\\[1mm]
\ds{2\expo{}a} & \ds{+} & \ds{\expo{}b} & \ds{=} &
\ds{-\,{2 \over 3}\,\expo{}}
\end{array}\right.
\\[2mm] &\ \implies a = -\,{3 \over 2}\ \mbox{and}\
b = {7 \over 3}
\\[5mm] &\
\bbx{\on{x}\pars{t} = {1 \over 6}\pars{t^{3} -
9t + 14}\expo{t}} \\ &
\end{align}

Title: Solving a 2nd order ODE


问:
$f(c) = \alpha c$ for $f:\left[0,1\right]\longrightarrow\left[0,1\right]$ continous function
find all values of $\alpha \in \mathbb{R}$ so for all $f:\left[0,1\right]\longrightarrow\left[0,1\right]$ that is continuous exists $c\in\left[0,1\right]$ that satisfies $f(c) = \alpha c$

Any ideas on how to find all of the possible values for $\alpha$?

答:
If $\alpha \geq 1$, then for every continuous function $f : [0,1]  \rightarrow [0,1]$, define $g(x)=f(x)-\alpha x$.
Then $g(0)=f(0)-\alpha\times 0 = f(0) \geq 0$ because $f$ takes its values in $[0,1]$. Similarly, $g(1)=f(1)-\alpha \times 1 \leq f(1)-1 \leq 0$ (because $\alpha \geq 1$ and $f(1)\leq1$). So because $g$ is continuous, the intermediate value theorem can be applied, so $g$ vanishes, ie there exists $c \in [0,1]$ such that $f(c)=\alpha c$.

If $\alpha < 1$, then for $f(x)=1$, there does not exists any $c$ such that $f(x)=\alpha c$.


答:
As with most definitions in distribution theory, it boils down to "formal integration by parts":

$$\langle \Delta u,\phi \rangle = -\langle \nabla u,\nabla \phi \rangle = \langle u,\Delta \phi \rangle.$$

In general when doing this you need to be careful to be sure that the boundary terms in each of these integrations actually vanish in your setting. In the case described in the question this is trivial since $\phi \in D(\Omega)$.

Title: Definition : The Distribution of laplacian of a function?

Body:
Let $u\in H_0^1(\Omega)$(but i think we can be way more general than than).

How is $\langle\Delta u,\phi\rangle$ (distribution) defined for $\phi \in D(\Omega)$ a test function ?


问:
What kind of a locus $Y$ describe when moving $\Omega $?
Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. Variable circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.

Suppose lines $AD$ and $BC$ meet at point $Y$.

What kind of a locus $Y$ describe when moving $\Omega $? Is that  some projective curve?

How can we recognize it with pure Euclidian geometry without any (or as few as possible) anaylitic calculation? For example, if we have a projective transformation $\pi$ between pencils of lines, then we know that $p\cap \pi(p)$ describes some conic or line.

答:
It's not clear how OP expects to "recognize [the locus] with pure Euclidian geometry". Cubic curves come in many shapes. So I cheated and used coordinates (and some symbol-crunching in Mathematica). Perhaps the finding here can inspire a "pure" solution.

Taking $T$ at the origin, $\angle ATB=2\theta$, $a:=|TA|$, and $b:=|TB|$, I get this equation for the locus:

$$\begin{align}
0 &= (x^2+y^2)(a+b) \cos\theta \left(\;x (a-b) \sin\theta - y (a + b) \cos\theta\;\right) \\[4pt]
&\quad+ 2 x^2  a b (2y -(a-b)\sin\theta )\\[4pt]
\end{align} \tag1$$
As a polar curve parameterized by $\phi$, this is:

$$r = \frac{2 a b (a - b)\sin\theta\cos^2\phi}{
  2 a b \sin2\phi\cos\phi
   - (a + b) \cos\theta (a \sin(\phi-\theta) + b \sin(\phi+\theta))
  )} \tag2$$

In barycentric coordinates $\alpha:\beta:\tau$ relative to $\triangle ABT$:

$$4 a b (a+b) \,\alpha\beta(\alpha - \beta) \sin^2\theta = (a - b) (a\alpha + b \beta)^2 \tau \tag{3}$$

Here are figures, with and without circle overlap:

Four points are common to all instances: $A$, $B$, $T$, and the midpoint of $\overline{AB}$ (labeled $M$).

When the circles overlap, their intersections ($P$ and $Q$) are on the curve.

The curve crosses the $x$-axis at $H := \left(\frac{2ab}{(a+b)\cos\theta},0\right)$. If $A_\star$ and $B_\star$ are the respective centers of the $A$ and $B$ circles, then $|TH|$ is the harmonic mean of $|TA_\star|$ and $|TB_\star|$.

Finding other landmarks and properties is left as an exercise to the reader.


问:
Why can't two planes be orthogonal in $R^3$?
I think my confusion comes from not knowing the difference between orthogonal and orthogonal subspaces.

So according to this, two planes can't be "orthogonal" in $R^3$. How does having more than three dimensions resolve this?

I think the argument is two planes can't be orthogonal subspaces to each other because their intersection is a line, so if we take any two vectors on that line dot them together, we won't have a zero dot product.

However, two planes can be geometrically orthogonal to each other right? For example, the xy and xz planes are perpendicular, so they must be orthogonal to each other right?

Can someone explain why having more than three dimensions means we can have orthogonal plane subspaces? Don't we still have the same problem as before?

答:
In the case of orthogonal planes, the requirement is that each plane
has a basis of two orthogonal vectors and that one of the vectors
of each pair is shared between the two planes and the other vector
of each pair form a pair orthogonal to each other.
In the case of orthogonal planes as subspaces, we require that each
vector of each pair is orthogonal to each vector of the other pair.
This requires the dimension of the ambient space to be at least $\,2+2=4.$


问:
Find the number of functions $f : [\![1,n]\!] \to [\![0,n]\!]$ such that $\forall k \in [\![1,n]\!], \exists k' \in \Bbb N, f^{k'}(k)=0 $.

Find the number of functions $f : [\![1,n]\!] \to [\![0,n]\!]$ such that $\forall k \in [\![1,n]\!], \exists k' \in \Bbb N, f^{k'}(k)=0$.

Of course here $f^{k'}(k)=f(f(\cdots(k)\cdots))$.

I've studied the cases $n=1$ ($f=0$ is the only solution) and $n=2$ ($3$ solutions). I had the feeling that there could be something related to disjoint cycles but this is not the case in general (if we take $f : (1,2,3) \mapsto (2,0,2)$ we have $f=(1 \to 2 \to 0)(3 \to 2 \to 0)$).

答:
Your interpretation of "disjoint cycles" are essentially paths that go from non-accessed nodes (the elements that are not in the image of your function) all the way down to $0.$ Show that these are, pictorially, rooted trees in the sense that there is a root(the number that goes to $0$) and there can not be cycles, otherwise, the elements in those cycles never reach $0$.

 This means that Cayley's theorem applies and there are $(n+1)^{n-1}$ of such functions. Convince yourself by drawing in this pictorial form the functions you got for $n=3.$ Notice that the $n+1$ comes from placing $0$ as the root of this tree (this is probably what Mike's hint was about).


问:
Analytical Solution of Heat Equation with Neumann Boundary Conditions
Consider the heat equation:

\begin{equation}
\frac{\partial u}{\partial t} = K \frac{\partial^{2} u}{\partial x^2}
\end{equation}

with initial condition:

$ u(x,0) = sin(\pi ^2 - x^2) $

and Neumann boundary conditions:

$ \frac{\partial u}{\partial x}(\pm \pi, t) = 0 $.

I have attempted using separation of variables, where:

\begin{equation}
u(x,t) = X(x)T(t)\\
\Rightarrow  \frac{\partial T}{\partial t} = -K\lambda T, \\
\Rightarrow  \frac{\partial^2 X}{\partial x^2} = -\lambda X
\end{equation}
and
$X'(-\pi) = X'(\pi) = 0$.

This leads to the solution of $T$:

\begin{equation}
T = A exp(-K \lambda t)
\end{equation}

However, I am stuck from this point onward in obtaining the solution for $X(x)$.

答:
As per Mark's hint, you should start with separating and solving for $X(x)$

Suppose $u(t,x) = T(t)X(x)$ and apply separation of variables:

\begin{align*} \frac{1}{K} \frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda \leq 0. \end{align*}

Solve for $X(x) = A\cos(\sqrt{\lambda}(x+\pi)) + B\sin(\sqrt{\lambda}(x+\pi))$ with boundary conditions $X'(\pm \pi)=0$ (and $X(x) \equiv C$ for $\lambda =0$). Argue that a non-trivial eigenfunction in terms of $\cosh$ and $\sinh$ cannot satisfy the boundary condition.

We must have $B=0$ as $X'(-\pi)=0$. Deduce $X'(\pi) = -2A\sqrt{\lambda}\sin(\sqrt{\lambda}2\pi) = 0 \implies \lambda = \left (\frac{n}{2} \right )^{2}$. So you will get eigenfunctions of form $X_{n}(x) = \cos \left (\frac{n}{2} (x+\pi) \right )$.

Solve for $T_{n}(t) = e^{-\left (\frac{n}{2}\right)^{2}Kt}$ and express as a sum

\begin{align*}
u(t,x) &= \sum_{n=0}^{\infty} C_{n}T_{n}(t)X_{n}(x) \\
&= \sum_{n=0}^{\infty} C_{n} e^{-n^{2}Kt/4} \cos \left (\frac{n}{2} (x+\pi) \right ).
\end{align*}

Solve for $C_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} u(0,x)\cos \left (\frac{n}{2} (x+\pi) \right )\, dx$ for $n \geq 1$ and $C_{0} = \frac{1}{2\pi} \int_{-\pi}^{\pi} u(0,x)\, dx$. These values may not be integrable in terms of elementary functions.


问:
Weak and weak$^*$-closure of convex set.
Let $H$ be a Hilbert space and $C \subseteq B(H)$ be a convex set. Is it well-known that the closure of $C$ with respect to the strong operator topology agrees with the closure of $C$ with respect to the weak operator topology.

Is it true that the closure of $C$ with respect to the weak operator topology also coincides with the closure of $C$ with respect to the $\sigma$-weak (= the weak $^*$-topology on $B(H)$, or equivalently, the topology generated  by the normal functionals) topology?

If the answer is positive, a reference suffices as an answer.

答:
Two locally convex topologies on a vector space share closed convex sets iff they have the same dual.

The dual of $B(H)$ with the weak topology is formed by the functionals
$$
  \phi_S: T\in B(H)\mapsto \text{tr}(TS)\in\mathbb C,
  $$
where $S$ is a finite rank operator, while the dual with the $\sigma$-weak topology is as above, but for the larger class of trace class operators $S$.

Therefore the answer to the question posed by the OP is no.

For a concrete example, given a trace class operator $S$ of infinite rank, the null space of $\phi_S$ is $\sigma$-weakly closed, but not  weakly closed.


问:
What is the value of $1^z$ for $z\in\mathbb{C}$?
According to Wolfram Alpha, $1^z=1$ for $z\in\mathbb{C}$. If this is true, then what is wrong with the following argument that $1^z$ has infinitely many values?

Let $z=x+iy$. Then,
\begin{align}
1^z &= 1^{x+iy} \\
&= 1^x \cdot 1^{iy} \\
&= 1^{iy} \\
&= e^{iy\log(1)} \\
\log(1) &= \{0,2i \pi, 4i \pi,6i \pi, \ldots\} \\
1^z &= \{e^{iy \cdot 0},e^{iy \cdot 2 i \pi},e^{iy \cdot 4 i \pi},e^{iy \cdot 6 i \pi},\ldots\} \\
&= \{e^0,e^{-2y\pi},e^{-4y\pi},e^{-6y\pi},\ldots\}
\end{align}

Only one of these values is equal to $1$.

答:
If, when $a>0$, you define $a^z$ as $\exp\bigl(z\log(a)\bigr)$, where $\log(a)$ is the only real logarithm of $a$, then indeed we always have $1^z=1$.

Otherwise, it's up to you to tell us how you are defining $1^z$. However, note that the set of all logarithms of $1$ is not $\{0,2\pi i,4\pi,i,6\pi i,\ldots\}$; it's $\{0,\pm2\pi i,\pm4\pi i,\pm6\pi i,\ldots\}$.


答:
I would suggest change of variable as -

$\displaystyle xy = u, \frac{y}{x^2} = v$

That leads to $\displaystyle J = \frac{1}{3v}$ and integrand $\displaystyle \frac{y^2}{x} = uv$

So the integral simplifies to $\displaystyle \int_1^4 \int_1^5 \frac{u}{3} du \, dv = 12$

The intuition for change of variable comes from seeing the curves they are bound by. As you have two curves with $xy = c$ and two curves with $y = cx^2$. So you can use change of variable to integrate over a rectangular region. One of them is $u = xy$ and the other can be $\frac{x^2}{y}$ or $\frac{y}{x^2}$. I chose the latter as it simplified the integrand as $\frac{y^2}{x} = uv$.

Title: Double integral over a typical region.

Body:
Question: Let $R$ be the region in $xy$-plane bounded by the curves $y=x^2,y=4x^2,xy=1,xy=5$.
Then find the value of line integral $\iint_R\frac{y^2}{x}dydx$.

I found it tough and time taking if I evaluate this integral from general method, is there any theorem which make it easy? Since the question is asked in competitive exam so I hope there must be an easy and fast way to solve it.

Any help will be appreciable.

Thanks in advance.


问:
Interchange order of integral
Can someone tell me which step in the process did wrong? My result seems wrong
$$D = \{(x, y) \mid \sqrt{2x - x^2} \leq y \leq \sqrt{2x}, 0 \leq x \leq 2 \}$$
$$\iint_D f(x,y) \, dy \, dx$$
Then I can draw the Graph

According to the graph, we can know that this is the area enclosed by $x = \frac{y^2}{2}$ and $x = 2$ minus the semicircle area
$$\iint_D f(x,y)\,dy\,dx = -\int_0^4 \int_{\frac{y^2}{2}}^2 f(x,y)\,dx\,dy - 2\int_0^1 \int_1^{1 + \sqrt{1 - y^2}} f(x,y)\,dx\,dy $$

答: You haven't said what function $f$ is, but your last equality is right if the integral of $f$ over the left half of the semicircle equals the integral over the right half. But not otherwise.

Title: Interchange order of integral


问:
Finding the equivalent of a polynomial sequence

Let $P_n=X^n+X-1$ be a polynomial with only one positive real root
$x_n$.

The sequence $(x_n)$ is obviously well defined and it's
not hard to find that $\lim\limits_{n \to +\infty} x_n=1$

In fact $(x_n)$ is an increasing bounded sequence. Let $l$ be its limit. We obviously have $l \le 1$.

As $P_n$ is increasing on $\mathbb{R_+}$,

$0=x_n^n+x_n-1\le l^n+l-1$ so the limit when $n \to \infty$ satisfies $l-1\ge 0$.

So $l=1$.

Then I need to find an equivalent sequence to $x_n-1$.

What I did:

We have $n=\frac{\ln(1-x_n)}{\ln x_n}$.

I therefore tried to find an equivalent of the form $n^{\alpha}(\ln n)^{\beta}$ and after some trials I found that $-\frac{\ln n}{n}$ works.

Since

$\displaystyle\frac{n}{\ln n} (1-x_n)=\displaystyle\frac{\frac{\ln(1-x_n)}{\ln x_n} (1-x_n)}{\ln\frac{\ln(1-x_n)}{\ln x_n}}$ we only need to show that $\displaystyle\lim\limits_{n \to +\infty}\frac{\ln(1-x_n)}{\ln\frac{\ln(1-x_n)}{\ln x_n}}=-1$

because $\lim\limits_{n \to +\infty}\frac{(1-x_n)}{\ln x_n}=-1$ (limit of the tangency point of $\ln$ at $1$).

we can show that $\displaystyle\lim\limits_{x \to 1^{-}}\frac{\ln(1-x)}{\ln\frac{\ln(1-x)}{\ln x}}=-1$

To simplify it, I noticed that both numerator and denominator tend to $\pm \infty$ when $x\to 1^-$.

So we can apply L'Hôpital's Rule and then :

$\frac{d}{dx} \ln(1-x)=\frac1{x-1}$

and $\frac{d}{dx}\left( \ln\frac{\ln(1-x)}{\ln x} \right)=\frac{x \ln x +(1-x)\ln{(1-x)}}{x(x-1) \ln x \ln(1-x)}$

the limit becomes $\frac{x \ln x +(1-x)\ln{(1-x)}}{x\ln x \ln(1-x)}=\frac 1 {\ln(1-x)}+\frac{(1-x)}{x\ln x} \to_{x \to 1^-} 0+(-1)=-1$.

Finally we have that $\frac{n}{\ln n} (1-x_n) \to -1$ and then $x_n-1 \sim -\frac{\ln n}{n}$

My questions:

Is this correct ?

Is there a better way to proceed ? Maybe with the MVT ?

Thanks in advance.

答:
I'll ignore the subscript for readability. Write $x = 1 - \frac{c}{n}$. We have $x^n = 1 - x$ which gives

$$\left( 1 - \frac{c}{n} \right)^n = \frac{c}{n}.$$

Taking logarithms gives

$$n \log \left( 1 - \frac{c}{n} \right) = \log c - \log n.$$

Approximating the LHS as $-c + O \left( \frac{c^2}{n} \right)$ gives $c + \log c = \log n + O \left( \frac{c^2}{n} \right)$ which gives $c \approx \log n - \log \log n$; estimating the error in this approximation involves estimating

$$\begin{eqnarray*} \log (\log n - \log \log n) &=& \log \left( \log n \left( 1 - \frac{\log \log n}{\log n} \right) \right) \\ &=& \log \log n + \log \left( 1 - \frac{\log \log n}{\log n} \right) \\ &=& \log \log n - \frac{\log \log n}{\log n} + O \left( \left( \frac{\log \log n}{\log n} \right)^2 \right) \end{eqnarray*}$$

which gives something like

$$c = \log n - \log \log n + \frac{\log \log n}{\log n} + O \left( \left( \frac{\log \log n}{\log n} \right)^2 \right).$$

Note that this error term is much larger than $O \left( \frac{c^2}{n} \right) = O \left( \frac{(\log n)^2}{n} \right)$; it decays very slowly. You can get more terms out of known asymptotics for the Lambert W function since we have $ce^c \approx n$ so $c \approx W(n)$.

To give a flavor of the accuracy here, when $n = 100$ we have (according to WolframAlpha)

$$x_{100} \approx 0.966584$$

and hence $c_{100} = 100(1 - x_{100}) \approx 3.342$. The approximation above for $c_{100}$ gives, taking one term at a time,

$$\log 100 \approx 4.605$$
$$\log 100 - \log \log 100 \approx 3.078$$
$$\log 100 - \log \log 100 + \frac{\log \log 100}{\log 100} \approx 3.410.$$

So taking the first term is not that great, and even taking the first three terms only gives us a single digit of accuracy. We also have $W(100) \approx 3.386$ which is at least two digits of accuracy. We could get more accuracy by starting with $W(n)$ and expanding from there, probably.


问:
If $f$ is a real-valued function such that the set $\{x : f(x)\ge r\}$ is Borel for any $r\in \mathbb{Q}$, then $f$ is a Borel function
I started with assuming $a$ and $b$ are rationals. Then I expressed an arbitrary interval $(a,b)$ as $$\left\{\bigcup_n \;\left[a+ \frac{1}{n},\infty\right)\right\} \; \bigcap \;  \left\{\bigcup_n \; \left[b+\frac{1}{n}, \infty\right)\right\}$$

It follows from the properties of the Borel sigma algebra and our given property that the above representation is a Borel set. But how can I extend this to a, b which are irrational?

I thought to use the the density of the rationals in the real numbers but I fail to see how I could apply this since $f^{-1}$ is not necessarily continuous so I can't just take the limit inside. Another tool which comes to mind is the fact that between any two real numbers there is a rational. But again I don't see how I can apply this.

答:
If $a$ is any real number, there is an increasing sequence of rational numbers $r_n$ such that $r_n< a$ and $r_n\to a$ as $n\to \infty$. Also,
$$\bigcap_{n}\{x|f(x)\geq r_n\}=\{x|f(x)\geq a\}$$

Hence the l.h.s is a countable intersection of Borel sets. Since $a$ was arbitrary, $f$ is Borel.


问:
Linear polynomials of finite fields
I have a final tomorrow, and I was looking over some exercises in my textbook. However, I can't seem to work this problem out.

Let $F$ be a field of $p^n$ elements and let $\alpha \in F^*$, where $F^*=F-\{0\}$.
Show $(x- \alpha)(x- \alpha^p)(x- \alpha^{p^2})\cdots(x- \alpha^{p^{n-1}}) \in \mathbb{Z}_p[x]$. Also, my text has some weird notation, $\mathbb{Z}_p$ is shorthand for $\mathbb{Z}/p\mathbb{Z}$.

It hints that I need to show both $\alpha+\alpha^{p}+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}\in\mathbb{Z}_p$ and $\alpha\alpha^{p}\alpha^{p^2}\cdots\alpha^{p^{n-1}}\in\mathbb{Z}_p$, but I'm having trouble with it. Help would be much appreciated.

答:
The easiest way to do this problem is the following: the map $\phi:\alpha \mapsto \alpha^p$ is an automorphism of $F$ whose fixed field is the prime subfield $\mathbb{Z}/p$. Furthermore, $\phi^n=1$ is the identity (since $F$ has $p^n$ elements). Your polynomial is therefore fixed by $\phi$, meaning that its coefficients are in $\mathbb{Z}/p$.

Here are the details if you want to do this from scratch (assuming Lagrange's theorem counts as "scratch"): first, for any $\alpha \in F$, we have $\phi^n(\alpha)=\alpha^{p^n}=\alpha$ by Lagrange's theorem applied to the group $F^\times$. This implies the fact mentioned above, that the coefficients of your polynomial are fixed by $\phi$.

In particular, taking $n=1$ shows that $\phi(\alpha)=\alpha$ for all $\alpha \in \mathbb{Z}/p$. In other words, every element of the prime field $\mathbb{Z}/p$ is a zero of the polynomial $x^p-x$. Now a polynomial of degree $p$ can have at most $p$ roots in any field, which shows that the prime subfield is exactly the set of $\alpha \in F$ with $\alpha=\phi(\alpha)$.

Title: Linear polynomials of finite fields


问:
linearity of deep learning model
So today I was watching a lecture that is taught by Ian Goodfellow about the adversarial attacks of the deep network. He made an interesting point I don't really understand. He said that for a deep neural network, since the activation function we mostly used is Relu, so the model becomes mostly linear. But the parameter of the network is not linear since the weight we multiply is nonlinear. I don't understand why he said that since when forward propagation we are going through those nonlinear calculations as well. so although the activation function is linear, when we used it to do the task, the model will be nonlinear as well?
The lecture link is right here https://www.youtube.com/watch?v=CIfsB_EYsVI&list=PL3FW7Lu3i5JvHM8ljYj-zLfQRF3EO8sYv&index=16

Anyone understand this and can clarify this for me? Thanks

答:
For most of the actual activation values at neurons, the neuron is operating in its linear range.

That's all it means.

Title: linearity of deep learning model


问:
Amy has a cuboid with volume at most 8

Amy has a clump of iron with volume $8$. She melts it and make a cuboid (she might have iron left over). Her boyfriend Ben increased the length, width and height of the cuboid by $1$ to form a new cuboid. Show that the surface area of the new cuboid is at least two times of the volume of the new cuboid.

I made this problem, so I know a solution (using inequality). Any solution is accepted as long as it makes sense. There might be a geometric solution that I don't know! I will post my solution if no one posts it.

答:
Let the sides be $a,b,c$, so $abc = 8$. We WTS

$$\sum 2(a+1)(b+1) \geq 2 \prod(a+1). $$

Dividing by 2 and expanding, we WTS

$$ a + b + c +3 \geq abc + 1 = 9$$

This is true because $ a + b + c \geq 3 \sqrt[3]{abc} = 6 $.

In the event that the volume is $ 0 \leq V \leq 8$, then use the fact that

 $$3\sqrt[3]{V} + 3 \geq  V + 1$$

If you're stuck, show your work and explain what you've tried.

Title: Amy has a cuboid with volume at most 8


问:
How to compute derivative when a function includes two indices
In a weighted graph, the probability of selecting an edge between nodes $i$ and $j$ is

$$P^{R}_{ij} := \frac{[a_{ij}]^\alpha [b_{ij}]^\beta}{\sum\limits_{r \in R} [a_{ir}]^\alpha [b_{ir}]^\beta},$$

where $a_{\cdot\cdot}$ is the weight of the edge corresponding to its argument nodes. $b$ is a heuristic function. $R$ is a subset of the graph's nodes, and $\alpha$ and $\beta$ are some positive real numbers.

In a design problem, I need to maximise $P^{R}_{ij}$ with respect to edge weights. However, it includes both $a_{ij}$ and $a_{ir}$ terms. Now, I am wondering whether I have to compute the total derivative taking $a_{ij}$ and $a_{ir}$ independent variables. So, what is that total derivative?

答:
$$
\frac{[a_{ij}]^\alpha [b_{ij}]^\beta}{\sum\limits_{r \in R} [a_{ir}]^\alpha [b_{ir}]^\beta}
$$
Consider the case where $|R|=3$ and $j=1.$ Then you have
$$
\frac{[a_{i1}]^\alpha [b_{i1}]^\beta}{\sum\limits_{r \in R} [a_{ir}]^\alpha [b_{ir}]^\beta} = \frac{[a_{i1}]^\alpha[b_{i1}]^\beta}{[a_{i1}]^\alpha [b_{i1}]^\beta+[a_{i2}]^\alpha [b_{i2}]^\beta+[a_{i3}]^\alpha [b_{i3}]^\beta} \tag 1
$$
You wrote:

I am wondering whether I have to compute the total derivative taking $a_{ij}$ and $a_{ir}$ independent variables.

Since $j=1$ in this case, that would mean $a_{i1}$ and $a_{ir}.$ But what is $r$? In the first term in the denominator, $r=1,$ so $a_{ir}= a_{i1}$ in the second term $r=2$ and $a_{ir} = a_{i2},$ and in the third term $r=3$ and $a_{ir} = a_{i3}.$


问:
How do linear transformation properties preserve vector space structure?
Currently I'm reading the Mathematics for Machine Learning book, and in the beginning of section 2.7 of the book, the following is given:

2.7 Linear Mappings

In the following, we will study mappings on vector spaces that preserve their structure, which will allow us to
deﬁne the concept of a coordinate. In the beginning of the chapter, we
said that vectors are objects that can be added together and
multiplied by a scalar, and the resulting object is still a vector. We
wish to preserve this property when applying the mapping:  Consider
two real vector spaces V,W. A mapping Φ : V → W preserves the
structure of the vector space if

Φ(x + y) = Φ(x) + Φ(y) (2.85)

Φ(λx) = λΦ(x) (2.86)

for all x, y ∈ V and λ ∈ R.

Generally, when vectors are added together, or multiplied by a scalar, the result is also a vector (the same type of object).

How are the above properties of a linear transformation preserving this?

As long as the LHS and RHS of both given properties belong to the same vector space, can I not consider the vector properties preserved? Why do they have to be equal?

What does it mean by they preserve the structure of a vector space?

Is there a particular reason Φ(λx + ψy) = λΦ(x) + ψΦ(y) has a form similar to a linear combination (y = a1.x1 + a2.x2 +...)?

答:
A mapping preserves some structure if the image of the mapping has the same structure as the domain. Examples include group and ring homomorphism, linear transformations and graph isomorphisms.

For linear algebra in particular we have a geometric notion that we can work that can give us a meaningful interpretation of this. Say we have a projection onto a plane in $\mathbb{R}^3$. This is linear transformation which means if we add the vectors, or scale the vectors before the projection it would be the same as scaling and adding them after the projection. It allows us to calculate in either space with the assurance that the image will be the same.


问:
Real Pole Residue theorem
I've been studying the residue theorem and I've been having a problem understanding the following result seen here (Eq.7.39) which states :

Given a regular function on the real axis, $g$, then $$\lim_{\epsilon \to 0}~\int_{-\infty}^{+\infty}\frac{g(x)}{x-x_0\mp i\epsilon}dx=PV\int_{-\infty}^{+\infty}\frac{g(x)}{x-x_0}dx \pm i\pi g(x_0)$$ where $x_0\in\mathbb R$.

What I don't understand is the factor $ i\pi g(x_0)$ instead of  $ 2i\pi g(x_0)$. Also, why $g(x_0)\equiv \operatorname{Res}(g(x),x=x_0)$ instead of $\operatorname{Res}(g(x),x=x_0\pm i\epsilon)$ in that same term?

Can anyone help me?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Usually
\begin{align}
\int_{-\infty}^{\infty}{\on{f}\pars{x} \over x \pm \ic \epsilon}
\,\dd x
& =
\int_{-\infty}^{\infty}\on{f}\pars{x}\,
{x \over x^{2} + \epsilon^{2}}\,\dd x\ \mp\
\ic\pi\int_{-\infty}^{\infty}\on{f}\pars{x}\,
{\epsilon/\pi \over x^{2} + \epsilon^{2}}\,\dd x
\end{align}
As $\ds{\epsilon \to 0^{+}}$, $\ds{{\epsilon/\pi \over x^{2} + \epsilon^{2}}}$ is a "Delta Dirac Representation" $\ds{\delta\pars{x}}$ and the RHS first integral becomes a Principal Value.
\begin{align}
\int_{-\infty}^{\infty}{\on{f}\pars{x} \over x \pm \ic \epsilon}
\,\dd x
& =
{\rm P.V.}\int_{-\infty}^{\infty}{\on{f}\pars{x}
\over x}\,\dd x\ \mp\
\int_{-\infty}^{\infty}\on{f}\pars{x}\,
\ic\pi\,\delta\pars{x}\dd x\ \mbox{as}\ \epsilon \to 0^{+}
\end{align}

Title: Real Pole Residue theorem


问:
Show that a multivariable function isn't injective anywhere

Let $f \in C^1(\mathbb{R}^n, \mathbb{R}^d), n>d $

Show that $f$ isn't injective in a neighborhood of $x \in \mathbb{R}^n, \forall x$

I thought about assuming there is such $x$ and thus a sphere $U=B_{\varepsilon}(x)$ where $f$ is injective, and then looking at $f:U \to f(U)$ which is invertible, trying to get to a contradiction somehow, but with no success so far.
Hints appreciated.

答:
Suppose it was injective in some open ball. By reducing the radius, we can assume it is continuous in a closed ball. Let $B$ be this ball.

We now restrict $f$ to $B$. Then we have a bijection $B \to f(B)$. This is a surjection from a compact space to a Hausdorff space, so it is a quotient map (it maps open sets to open sets). Hence, $f^{-1}$ is continuous on $f(B)$, so $f$ restricted to $B$ is a homeomorphism.

Note that $f(B)$ is compact and connected. Then, it is connected and bounded, so it is homeomorphic to a ball in $\mathbb{R}^n$. However, a ball in $\mathbb{R}^n$ is not homeomorphic to a ball in $\mathbb{R}^d$ if $n \neq d$.