问:
Proof regarding a function that is continuous and discontinuous on different domains.

I'm looking for feedback on my proof, as I am sure that there is something I've missed that can render the entire thing false. The whole topic of discontinuity and stuff I can and cannot use is really confusing to me and I'd love to hear some tips for my future proofs in general.. Of course if there is something I did wrong, or is inaccurate in my proof I'd love to hear it!

Note that D(x) is the Dirichlet function.

Consider the function $f: R➜R$,    $\ f\left(x\right)\ =\ \sin\pi x\  \cdot\ D\left(x\right)$.

First we prove f is continuous at every $x ∈ Z$. Note that $\sin\pi x\  = 0$ for every $x ∈ Z$, and is continuous at every $x ∈ R$.

Let $x_{0}$ ∈ Z. Since D(x) is bounded, and since

$\lim \:_{x\to \:x_{0}}\sin \pi \:x=\lim \:\:_{x\to \:\:x_{0}}\sin \:\pi  \:\:x_{0}=0$

Then by "Bounded x Vanishing" we have

$\lim _{x\to x_{0}}\left(\sin \left(\pi \:x\right)\:\cdot  D\left(x\right)\right)=\left(\sin \:\left(\pi \:\:x_{0}\right)\:\cdot  \:D\left(x_{0}\right)\right)=0\ \cdot\ 1 = 0$

Therefore f is continuous at every $x ∈ Z$

Now we prove f is discontinuous at every $x ∈ R/Z$.

Note that $\sin\pi x\ ≠ 0$ for every $x ∈ R/Z$.

We have previously proven that a function of the sort $g: R➜R$, $\  g=f\cdot D$ is continuous at $x_{0}$ if and only if $f$ is continuous at $x_{0}$ and
$f(x_{0}) = 0$.

Since in our case we can replace $f$ with $\sin\pi x\ $, as it is
continuous on every $x ∈ R$, and we know that $\sin\pi x\ ≠ 0$ for
every $x ∈ R/Z$, then our function cannot be continuous at any $x_{0} ∈  R/Z$, and therefore is discontinuous at every $x_{0} ∈ R/Z$.

答:
What you did looks correct.

However the "most difficult" part of the proof lies in the sentence We have previously proven that a function of the sort $g: \mathbb R \to \mathbb R$, $\
 g=f\cdot D$ is continuous on $a$ if and only if $f$ is continuous and
$f(a) = 0$ which is not provided here.

And by the way a better sentence would be a real $g: \mathbb R \to \mathbb R$, $\
 g=f\cdot D$ is continuous at $a$ if $f$ is continuous at $a$ and
$f(a) = 0$. The continuity of $f$ for $b \neq a$ is irrelevant for the continuity of $g$ at $a$.


问:
Using Gauss-Bonnet to prove that geodesics have at most one point of intersection
Given an oriented Riemannian manifold $(M,g)$ of dimension $2$, such that $M$ has negative Gaussian curvature everywhere and $M$ is diffeomorphic to $\mathbb R^2$, I'm looking for a way to show that two geodesics are either disjoint, coincide or intersect in precisely one point, using the Gauss-Bonnet theorem.

Lacking any insight, I'm trying to find a way by contradiction: If there are two distinct geodesics intersecting in two points $p$, $q$, then we obtain a simple curve in $M$, consisting of two geodesic segments, by first following along one geodesic from $p$ to $q$ and then following along the other geodesic back from $q$ to $p$. Since $M\cong \mathbb R^2$, this simple closed curve bounds a simply connected domain $\Omega \subset M$ ($\Omega$ is diffeomorphic to a disk).

Let $\theta_1, \theta_2$ be the exterior angles between the two geodesic segments at the end-points $p$ and $q$. The Gauss-Bonnet theorem applied to this situation yields
$$\int_\Omega K + (\theta_1 + \theta_2) = 2\pi \chi(\Omega) = 2\pi.$$
(where the geodesic curvature parts drops out, because we have two geodesic segments). Since $K<0$, this implies that $\theta_1 + \theta_2 > 2\pi$.

Can anyone see how to finish the argument from here? Or maybe someone has a better idea?

Thanks for your help!

答: In general, at a corner, we have  $|\theta_j|\le \pi$. Here we have $|\theta_j|<\pi$ because these are geodesic segments, so $\theta_1+\theta_2<2\pi$. Sam, your argument is good. </p>

ContentLicense: CC BY-SA 3.0


问:
Is this proof for countability correct?
Conjecture: The set A = {(a1, a2, a3,..)} of infinite sequence of integers is  countably infinite.

I think that this is true (correct me if I'm wrong) but I'm not sure if my proof is correct. Can someone help?

Here's my proof:

Firstly, we agree to list the sequence in an increasing order, i.e. a1≤a2≤a3.... so on.
Then we can list the elements of A as follows:

Compare the first entry in the tuples(a1). Then arrange the tuples in an increasing order of them. Repeat the same for the second entry and so on...
By the end, we would have a list of the elements of A. Thus it is countably infinite.

I highly doubt the validity of this "sorting the tuples" proof. If it's wrong, please also supply a correct proof.
Thanks.

答:
Your claim is false. There are two fairly instructive ways to show this. It is perhaps easier to think of $A$ as the set of functions $f:\mathbb N\to\mathbb Z$.

First, $\mathcal P(\mathbb N)$ bijects naturally with the set of functions $\mathbb N\to\{0,1\}$ by mapping a set $S\in\mathcal P(\mathbb N)$ to its indicator function $\mathbf 1_S(n)=\begin{cases}1\text{ if }n\in S\\0\text{ if }n\not\in S\end{cases}$. So $\mathcal P(\mathbb N)$ injects into $A$, hence $A$ is uncountable.

Alternatively, we can use Cantor's diagonal argument. Suppose $A$ is countable, list it as $f_1,f_2,\dots$. Then define $f(n)=\begin{cases}1\text{ if }f_n(n)\neq 1\\2\text{ if }f_n(n)=1\end{cases}$. But $f$ is not on the list of $f_1,f_2,\dots$, contradiction.

What is true though is that the set $B$ of functions $\mathbb N\to\mathbb Z$ with finite support is countable (i.e. in your original notation, the set of integer sequences that are eventually constantly $0$).

ContentLicense: CC BY-SA 4.0

Title: Is this proof for countability correct?


问:
Product and Chain Rule(conceptual question)
Right from 10th, my teachers gave me all the formulas and techniques on how to solve problems in calculus.

Product rule= $$\frac{d}{dx}f(x)g(x)= f(x)g'(x)+g(x)f'(x)$$

Chain rule= $$\frac{dy}{dx}=\frac{dy}{du}* \frac{du}{dx}$$

Why is it that while applying product rule, one function is kept constant, while the other differentiated?

I saw the video on the visual representation of the chain rule and product rule on 3Blue1Brown, yet I'm unable to formulate an explanation for it. Especially when we increase the number of functions and variables.

I would really appreciate any help regarding this. Thanks in advance!!

答:
I like to think about how the area of a rectangle changes when we vary the sides:

$$(a+\delta_a)(b+\delta_b)-ab=\delta_a b+a\delta_b+\text{ something too small to worry about.}$$


问:
Trying to prove the inequality $P(h(X)≥a)≥(E[h(X)]−a)/(α − a)$ in probability and measure theory
I've been struggling for hours. The full details of the question is,

For a function $h:\mathbb{R}\rightarrow[0,\alpha]$ that is nonnegative and bounded and $0\leq a<\alpha$ show that,

$P(h(X)\geq a)\geq \frac{E\{h(x)\}-a}{\alpha-a}$

I'm currently doing my bachelor in mathematics and I'm halfway through a integral- and measure theory course and I'm just going through old assignments seeing if I can solve them during the christmas break.

What I have tried to do is find something similar to $P(h(X)\geq a)$ and using the theorem that $P(h(x)\geq a)\leq\frac{E\{h(x)\}}{a}$. Further I would apply that $P(A)=1-P(A^{\textbf{c}})$, so I could flip the inequality around when multiplying by one.

答:
Hint: Observe that $0 \le h(X) \le \alpha$, so for any $a \in [0,\alpha)$, $-a \le h(X) - a \le \alpha - a$.  Use the monotonicity of the expectation to see that $E[h(X)] - a \le \alpha - a$.

What's missing is $P(h(X) \ge a) = E[1_{\{h(X) \ge a\}}]$.  Can you put this back to the previous inequality?

 In the first inequality, multiply both sides by the above indicator function to get  \begin{align} h(X) - a \le& (h(X) - a) 1_{\{h(X) \ge a\}} \le (\alpha - a) 1_{\{h(X) \ge a\}}. \end{align}  Then take the expectation to conclude.


问:
Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$
This problem $5b$ from Chapter $14$ of Spivak's Calculus:

Find a function $g$ such that $\int_0^{x^2}tg(t)dt = x + x^2$ (Note $g$ is not assumed to be continuous at $0$).

I think this question is impossible? First of all there's no restrictions on $x$, so letting $x = 1$ and $x = -1$, we get $0 = 2$.

So maybe we have to restrict to $x \geq 0$? In that case, we can take the square root:

Find $g$ such that $\forall x \geq 0 : \int_0^{x}tg(t)dt = x + \sqrt{x}$

Let $f(t) = tg(t)$, so we need $\int_0^{x}f(t)dt = x + \sqrt{x}$

Now $\frac{d}{dx} (x + \sqrt{x}) = 1 + \frac{1}{2\sqrt{x}}$. But $f(t) = 1 + \frac{1}{2\sqrt{t}}$ won't work because $1 + \frac{1}{2\sqrt{t}}$ isn't bounded as $t \rightarrow 0$, so the integral won't be defined.

Am I missing something?

答:
The function $$g(x) = \frac{1 + 2\sqrt x}{2(\sqrt x)^3}$$ seems to satisfy the condition. Below is how to find it:

Let $G(x) = \int xg(x)dx$. Then, we have
$$\int_0^{x^2}tg(t)dt \overset{\text{(FTC)}}= G(x^2) - G(0) \overset{*}= x + x^2
 $$
Now, differentiate both sides of $(*)$ to get
$$\begin{align} 2x[x^2g(x^2)] &= 1 + 2x \\[1mm] 2x^3g(x^2) &= 1+ 2x \\ g(x^2) &= \frac{1+2x}{2x^3}.\end{align}$$ Therefore, we get $$g(x) = \frac{1 + 2\sqrt x}{2(\sqrt x)^3} $$

Indeed,
$$\int_0^{x^2} t \frac{1 + 2\sqrt t}{2(\sqrt t)^3} dt = \int_0^{x^2}\left( 1 + \frac{1}{2\sqrt t}\right)dt = \left[\sqrt t + t\right]_0^{x^2} = x + x^2$$ as desired.


问:
Prove that $\triangle AMC$ is an isosceles right triangle

In $\triangle ABC$ construct squares $ABDE$ and $BCFG$. Let $M$ be the midpoint of $EF$ then, prove that $\triangle AMC$ is an isosceles right triangle.

I was able to prove that $\triangle AMC$ is an isosceles only and here is my solution:

The area of $\triangle BAF$ is equal to the area of $\triangle EBC$ so we get
\begin{equation}
AE×AF\sin \angle BAF=CF×CE\sin \angle ECB
\end{equation}
And using law of cosines on $EF$ we get
\begin{equation}
AE^2+AF^2=CF^2+CE^2
\end{equation}
By the length formula of median $AM$ and $CM$ we get that $AM=CM$. I know this solution is kinda rush and not give much detail but I'm not really good at Latex, please don't mind me. Please help prove that it is the right triangle. Thanks!

BTW, $N$ in the photo is the point where $AE$ meet $CF$

答:
Let $P$ and $Q$ be the midpoints of $BF$ and $BE$ respectively. Draw $MQ$, $MP$, $AQ$ and $CP$.

Now, notice that, $\frac{AQ}{AB}=\frac{QM}{BC}=\frac{1}{\sqrt2}$ and  $\angle AQM=\angle EQM-90=\angle EBF-90=\angle ABC$; Hence $\triangle AQM\sim \triangle ABC$.

Thus, $\angle MAC=\angle QAB=45^{\circ}$
[ Subtract $\angle BAM$ from both $\angle BAC$ and $\angle QAM$ ]

In a similar way, prove that $\triangle MPC\sim \triangle ABC$. Then, $\angle ACM=\angle BCP=45^{\circ}$.

In $\triangle AMC$, $\angle MAC=\angle ACM=45^{\circ}$; Therefore, it is an isosceles right triangle.


问:
$\mathbb{R}$ is closed under division
The set of real numbers $\mathbb{R}$ is closed under division. Does that mean $0$ is also considered?

more specifically, should it be $\mathbb{R}-\{ 0 \}$?

because division by $0$ is not defined.

答: The real numbers aren't closed under division unless you don't include zero. So it is like you said: $\mathbb{R}-\{0\}$ is closed under division since we don't encounter that problem.

Title: $\mathbb{R}$ is closed under division


问:
How many passwords can we create that contain at least one capital letter, a small letter and one digit?
I have a problem in combinatorics.
How many passwords can we create such as:

Each password has length $n$ when $n\ge3$.

Each password must contain at least one capital letter (there are 26 letters in English), and at least one small letter, and at least one digit (there are 10 possible digits)

I tried solving this problem like this:

There are $62^{n}$ possible passwords without any restrictions

There are $2*26^{n}$ passwords with only capital letters or small letters.

There are $10^{n}$ passwords with only digits.

There are $52^{n}$ passwords with only capital letters and small letters.

There are $36^{n}-10^{n}-26^{n}$ passwords with only capital letters and digits.

There are $36^{n}-10^{n}-26^{n}$ passwords with only small letters and digits.

So to get the "right answer" I subtracted all of them from $62^{n}$ and got $62^{n}-52^{n}-2*36^{n}+10^{n}$

Is this the right method/answer? or did I miss something silly?
I am new to combinatorics and I want to make sure, I am on the right track.
Thanks in advance guys!

答:
Using Inclusion-exclusion principle, the answer is $62^{n}-52^{n}-2 \times 36^{n}+10^{n} + 2 \times 26^{n}$. Consider different cases and you will simply get the answer.

More precisely, first consider the passwords with no capital letter, then with no small letter and at last with no digit. Now, add the passwords with no capital letters and small letters (equivalently with only digits), then add passwords with no capital letters and no digits (equivalently with only small letters) and at last, add passwords with no small letters and no digits (equivalently with only capital letters).


答:
$52^n$ passwords with only capital letters and small letters also count where you have only small letters or capital letters. Like in other cases, you should have subtracted $2 \times 26^n$ from it. That is the miss in your solution otherwise it is correct. Here is how I would do it -

If $A, B, C$ are sets of passwords where we have capital letters, small letters and numbers missing respectively,

$|A| = |B|= 36^n, |C| = 52^n$

$|A \cap B| = 10^n, |B \cap C| = 26^n, |A \cap C| = 26^n$

$|A \cap B \cap C| = 0$

$|A \cup B \cup C|$ will give you all arrangements where one or more of the three are missing.

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|$

So, $|A \cup B \cup C| = 2 \times 36^n + 52^n - 2 \times 26^n - 10^n$

Answer you are interested in $ = 62^n - |A \cup B \cup C| = 62^n - 52^n - 2 \times 36^n + 2 \times 26^n + 10^n$

Title: How many passwords can we create that contain at least one capital letter, a small letter and one digit?

Body:
I have a problem in combinatorics.
How many passwords can we create such as:

Each password has length $n$ when $n\ge3$.

Each password must contain at least one capital letter (there are 26 letters in English), and at least one small letter, and at least one digit (there are 10 possible digits)

I tried solving this problem like this:

There are $62^{n}$ possible passwords without any restrictions

There are $2*26^{n}$ passwords with only capital letters or small letters.

There are $10^{n}$ passwords with only digits.

There are $52^{n}$ passwords with only capital letters and small letters.

There are $36^{n}-10^{n}-26^{n}$ passwords with only capital letters and digits.

There are $36^{n}-10^{n}-26^{n}$ passwords with only small letters and digits.

So to get the "right answer" I subtracted all of them from $62^{n}$ and got $62^{n}-52^{n}-2*36^{n}+10^{n}$

Is this the right method/answer? or did I miss something silly?
I am new to combinatorics and I want to make sure, I am on the right track.
Thanks in advance guys!


问:
Convergence of $\int_0^1 (\log \frac 1 x)^n dx$
As far as I can tell, the only issue is at $x = 0$ (since the function goes to infinity). However, if I take a function $x^{-p}, 0<p<1$, then $f(x)/g(x) \rightarrow 0$ as $x \rightarrow 0$.
Therefore, by ratio test, the integral should converge for all $n$ (since $g(x)$ is convergent).

However, the answer says that the integral is convergent only for $0 > n > -1$. Any hints would be very appreciated.

答:
The correct answer should be that the integral only diverges for $n\le-1$, and converges elsewhere. In fact, for any $n>-1$, the integral is not only convergent but also has a name - the Gamma function.

We will discuss each case:

1). For $n\le-1$, the integrand is bounded (tends to $0$ actually) near $x=0$. Near $x=1$, we have
$$-\log x = (1-x)+o(x-1)$$
and therefore $(-\log x)^n\sim (1-x)^n$, clearly unintegrable for $n\le-1$.

2). For $n\in(-1,0)$, the integrand is bounded (tends to $0$ actually) near $x=0$. Near $x=1$, we have
$$-\log x = (1-x)+o(x-1)$$
and therefore $(-\log x)^n\sim (1-x)^n$, clearly integrable for $n\in(-1,0)$.

3). For $n > 0$, the integrand is bounded (tends to $0$ actually) near $x=1$. And near $x=0$, we have
$$-\log x = \log\left(\frac1x\right)< \left(\frac1x\right)^\alpha,\forall \alpha >0$$
Take, for example, $\alpha$ to be $1/(2n)$. Since $x^{-\frac12}$ is integrable near $0$, so must be $(-\log x)^n$.

4). The case $n=0$ is trivial.

Evaluating the integral

Let
$$I_n(a,b):=\int_a^b(-\log x)^n dx,\quad n>-1,\,\, 0<a\le b < 1$$

Note
$$\frac{d}{dx}(x(-\log x)^n)=(-\log x)^n - n (-\log x)^{n-1}$$
whence
$$I_n(a,b)-nI_{n-1}(a,b) = b(-\log b)^n - a(-\log a)^n$$

Define
$$I_n(0,1):=\lim_{\epsilon\to 0+}I_n(\epsilon,1-\epsilon)=\int_0^1(-\log x)^n dx$$. Then
$$I_n(0,1)-nI_{n-1}(0,1)=0.$$

And
$$I_n(0,1)=\left(\prod_{k\in A}k\right)I_\gamma(0,1)$$
where
$$A:=\{k\in n+\Bbb Z\mid 0<k\le n\},\quad \gamma = \min\{k\in n+\Bbb Z\mid k>-1\}$$

When $n\in\Bbb Z_{\ge 0}$, we obtain the fammiliar result
$$I_n(0,1)=n!$$


问:
Prove $2^{123}<5^{53}$ without calculator.

Show in an algebraic way, not using calculator if possible that: $$2^{123}<5^{53}$$

First I want to tell that I'm trying to make this post a reference post.

There are duplicates out there, but none really offers a non-calculator proof, at most they rely on a fine approximation of $\log_{10}(2)$, and moreover they are all getting deleted one by one due to no effort showing from the original poster.

Since I found a solution, I'd like to expose it here, of course other solutions are welcome as well.

Secondly, the problem is not as easy as it looks and I'll show why:

The standard idea would be to compare $2^7\approx 5^3$ ($128$ vs $125$) but this inequality is the wrong order $5^3 < 2^7$.

By using it we can prove that $\ 2^{123}>16\cdot 5^{51}\ $ while we want in fact $\ 2^{123}<25\cdot 5^{51}$

There is not much room between $16$ and $25$ so the inequality is quite tight.

Another choice would be to consider other approximations like $\ 2^{16}<5^7\ $ or $\ 2^{30}<5^{13}$

The latter one is already quite tedious to calculate manually, yet it is not sufficiently tight:

You get $2^{123}=2^{30\times 4+3}<8\times 5^{13\times 4}<8\times 5^{52}$ but still $8>5$ and cannot conclude.

Note:

Some other posters have shown that using $\log_{10}(2)\approx 0.3010 < 0.3011$ was adequate

to prove the claim since $176\times 0.3011 < 53$.

答:
No logarithms around in this solution, and a calculator need not be at hand.

Let's start by sharpening the claimed bound: consider
$$1 \;<\;\frac{5^3}{2^7}\cdot\frac{5^{53}}{2^{123}} \;<\;\frac{5^{53}}{2^{123}}$$
then focus on the left "$<$" whilst the right one follows from $\,5^3=125 <128=2^7$.

The cube $\,63^3=250\,047\,$ is just a few bits away from $\,250\,000$, thus
$$1.26^3 \;=\;\left(\frac{63}{50}\right)^3 \;=\;\frac{250047}{125000}
\;\gtrsim\;\frac{250000}{125000} \;=\;2$$
so $1.26$ is slightly above $\sqrt[3\,]{2}\,$. That yields
$$\frac{5^{56}}{2^{130}} \;=\;\frac{5^{54}}{2^{108}\cdot 2^{18}}\cdot\frac{5^2}{2^4}
\;>\;\left(\frac{1.25}{1.26}\right)^{54}\cdot\left(\frac{5}{4}\right)^2$$
and the claimed bound is established if
$\left(\frac{1.25}{1.26}\right)^{54} =\left(1- \frac1{126}\right)^{54} >\big(\frac{4}{5}\big)^2\,$
is shown.
By truncating the Binomial expansion after four terms and upon simplification (without calculator $\ddot\smile$) we obtain
$$\begin{align}\left(1- \frac1{126}\right)^{54}
 & \;>\;1-\frac{54}{126} \;\;+\;\frac{54\cdot 53}{2}\cdot\frac1{126^2}
 \,-\,\frac{54\cdot 53\cdot 52}{2\cdot 3}\cdot\frac1{126^3} \\[2ex]
 & \;=\quad\frac8{14} \quad +\;\frac{27\cdot 53}{126^2} \left(1-\frac{52}{3\cdot 126}\right) \\[2ex]
 & \;=\quad\frac8{14} \quad +\;\frac{53\cdot 163}{7\cdot 126^2} \\[2ex]
& \;>\quad\frac8{14} \quad +\;\frac{49\cdot 162}{7\cdot 7^2\cdot 18^2} \;=\;\frac9{14}
\end{align}$$
The truncation is okay as the Binomial expansion has alternating signs and each plus-minus-pair is positive. Because of $\,\frac9{14} - \frac{16}{25} = \frac1{14\cdot 25}\,$ we are done.

In summary it has been shown that
$$\frac{5^{53}}{2^{123}} \;>\;1.024\,=\,\frac{2^7}{5^3}\,.$$


问:
Hyperbolic functions simplifying
How do you simplify

$$\cosh(\sinh^{-1}(x))$$

to become

$$(1+x^2)^{1/2}$$

I have managed to get $(1+\sinh^2(\sinh^{-1}(x))^{1/2}$ but haven't been able to progress from there.

答:
From $$\cosh^2t-\sinh^2t=1,$$ draw

$$\cosh t=\sqrt{1+\sinh^2t}$$ and set $$t=\sinh^{-1}x.$$

Title: Hyperbolic functions simplifying


答:
By definition you have
$$cosh(x)=\frac{e^x+e^{-x}}{2}$$
and
$$ sinh(x)=\frac{e^x-e^{-x}}{2}$$
You can compute the inverse function and find
$$sinh^{-1}(x)=ln(x+\sqrt{x^2+1})$$
Which substituted in your expression gives
$$\frac{ e^{ln{(x+\sqrt{x^2+1})}}+e^{-ln(x+\sqrt{x^2+1})}}{2}$$

Now remember that logarithm is the inverse of exponential, notice that $\frac{1}{a+b}=\frac{1}{a+b}\cdot \frac{a-b}{a-b}$ and $-log(x)=log(\frac{1}{x})$. Your result will follow.

Title: Hyperbolic functions simplifying

Body:
How do you simplify

$$\cosh(\sinh^{-1}(x))$$

to become

$$(1+x^2)^{1/2}$$

I have managed to get $(1+\sinh^2(\sinh^{-1}(x))^{1/2}$ but haven't been able to progress from there.


问:
Generators of the conformal group in 1 dimension
I was studying this book about the conformal group, and it explains the generators of the group in this way: 1 2

I don’t get why you can’t obtain the generators in d=1 following this reasoning. Everything there is also valid in d=1 except the L generator is zero. How to generalize this reasoning in d=1?

For example in this paper https://inspirehep.net/literature/108211 there are the generators in d=1.

The book is Conformal Field Theory by Philippe di Francesco.

SOLVED: The solution was explained later in the book for the case d=2 and the reasoning is the same as in the case d=1: 3

答:
Notice that if d=1 the conformal map equation does not put any constraint on the scale factor f(x), any smooth and invertible map is conformal. When d=1, every generator vanishes except translation, CST, and dilation. One example of such a conformal theory is conformal quantum mechanics, symmetric under SL(2,R), see:

https://arxiv.org/abs/1506.05596

You don't have to generalize the reasoning in your book, as the discussion is perfectly valid when d=1. (You get three generators, one of which is the Hamiltonian of the system, the other two are just "t" times the preceding generator, you can check the commutators and confirm that this is indeed a SL(2,R) algebra)


问:
Compact hyperbolic three-manifold - a question
In a recent paper the authors considered a spacetime described by

$$AdS_4 \times \Sigma_3 \times \mathcal{I}_r \times S^2$$

where $\mathcal{I}_r$ is an interval of the $r$-coordinate and the two-sphere $S^2$ is fibered around the manifold $\Sigma_3$. For consistency reasons of the construction, the manifold $\Sigma_3$ has to be a compact hyperbolic three-manifold. However they do not give the metric of that subspace. I was wondering how I could write the invariant line element of a compact hyperbolic three-manifold. That is, how to write

$\begin{equation}
ds^2_{\Sigma_3} = g_{ab} dx^a dx^b = \cdots
\end{equation}$

explicitly.

In a later section of the paper they also say that they fix the Ricci scalar of $\Sigma_3$ to be equal to minus six; $R=-6$. This is another question that I have; does this uniquely determine which compact hyperbolic thee manifold $\Sigma_3$ has to be?

Any explicit answers or suggestions to literature would be greatly appreciated. Please keep in mind that while I do have some maths background, my main discipline is theoretical physics.

答:
(This is packaged together from a series of comments.)

Hyperbolic manifolds of a fixed dimension are locally identical (not just topologically, which is true for all manifolds, but metrically). If you want a good local picture of 3-manifolds, look at the Poincaré half-space, which has a very simple model. The space is $\{ (x,y,z) \mid z > 0\}$, and the metric is:
$$ (ds)^2 = \frac{(dx)^2 + (dy)^2 + (dz)^2}{z^2}.$$

Intuitively, as one gets closer to the equator $\{z = 0\}$ distances grow. A nice way of thinking about hyperbolic geometry is Google Maps. Supposing the earth is flat (as some believe), let the $z$ coordinate be distance from the ground. The metric will then be the amount of time it takes to scroll from one point to another. As anyone who has used Google Maps can confirm, the fastest way from Marrakesh to Maine is to zoom out, pan across, and zoom back in. This is what hyperbolic geodesics look like too: distances are huge close up, so the fastest path is to back up and then re-approach.

Since the geometry is the same everywhere, what differentiates different hyperbolic manifolds is how they are glued together. For $n=2$, there is a rich gluing theory, called Teichmüller theory. For $n\geq 3$, Mostow's rigidity theorem tells us that topology determines geometry (for complete, finite-volume manifolds, and hence, in particular, compact ones).

A nice large-scale geometric property of hyperbolic manifolds is that their geodesic flow is ergodic, so if you start at a random point and start walking in a random direction, you'll very likely visit almost all of the space in a statistically nice way. The "universal" half-space model of Hyperbolic space has the feature that if you and a friend start walking from the same point in different directions, you end up far apart much more quickly than in flat, Euclidean space (this is the "exponential divergence" of geodesics in hyperbolic space). Said another way, the volume of balls in hyperbolic space grow exponentially, although this is only true locally in any given compact space, where the volume is ultimately bounded.

In general, visualizing a particular hyperbolic 3-manifold is tough. There are, however, a variety of concrete constructions that give more intuition. You can build hyperbolic 3-manifolds out of (a pair of) hyperbolic surfaces, you can build a hyperbolic 3-manifold by crossing a surface with an interval and gluing the ends with a very fancy twist (actually, a random twist works most of the time!), you can build a hyperbolic 3-manifold by cutting a link out of the $3$-dimensional sphere and gluing it back with a flourish (Dehn surgery, Lickorish-Wallace theorem), you can build a hyperbolic $3$-manifold by taking limits of the objects described above, and there are other combinatorial and algebraic constructions that I find a little harder to picture.


问:
Basis for Eigenspaces
Consider the real matrix $\alpha = \begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}$ and let $\beta \in M_n (\mathbb{C})$ be the same matrix but considered as a complex matrix (so over the complex numbers)

Find the basis of the corresponding Eigenspace

I found found the eigenvalues to be:

$\alpha$: over reals and then only the value $\lambda_1=3$

$\beta$: over complex and then the values $\lambda_1=3$,  $\lambda_2=i$ and  $\lambda_3=-i$

How would I proceed to find a basis for the Eigenspaces of the two matrices$

答:
For example, we can find the eigenvector corresponding to $\lambda = i$ as follows: compute
$$
\beta - \lambda I =
\pmatrix{
7-i &3  &-4 \\
 -2&-1-i  &2 \\
 6&2  &-3-i }.
$$
We can row-reduce this matrix to get the reduced row-echelon form
$$
\pmatrix{1 & 0 & -1/2 - i/2\\0 & 1 & i\\ 0 & 0 & 0}.
$$
Conclude that the kernel of $A - \lambda I$ (i.e. the eigenspace) is spanned by the vector (i.e. eigenvector) $v_i = (1/2 + i/2, -i, 1)^\top$. In other words, the set $\{(1/2+i/2,-i,1)^\top\}$ forms a basis of the eigenspace associated with $\lambda = i$.

The other two basis (each a set with one vector) can be computed in a similar fashion.

Actually, because $A$ has real entries, we can use our result for $\lambda = i$ to get the eigenvector for $\lambda = -i$:
$$
Av_i = iv_i \implies \overline{Av_i} = \overline{iv_i} \implies A \overline{v_i} = -i \overline{v_i}.
$$
So, we can take $v_{-i} = \overline{v_i} =(1/2 - i/2, i,1)^\top$.

Title: Basis for Eigenspaces


问:
Primes in the form $n^{k}+n-1$
Lately I've been (very casually) toying with primes in the form $n^k+n-1$, as a very far-reaching generalization of Fermat primes. (you get a Fermat prime when you set $k=2^m$ and $n=2$). I have little training in number theory, so mainly I've been just typing stuff into Wolfram and searching for patterns.

One thing I started doing was looking for the smallest $k$ such that $n^k+n-1$ is prime for any particular $n$. I've been able to find solutions for $n \leq 106$. Obviously $k=1$ and $k=2$ do a lot of heavy lifting, but there are some remarkable numbers here. The one that stunned me the most upon finding it was $32^{108}+31 \approx 3.6 \times 10^{162}$, but there's an even bigger smallest solution: $80^{194}+79 \approx 1.58 \times 10^{369}$.

For $n=107$, there are no solutions for $k \leq 495$, which is about as far as I can reliably check.  Obviously any $k$ that works must be divisible by $4$.

So, three things I wonder about are:

Is any solution for $n=107$ known? Does it exist for sure?

In general, has it been shown that there is a solution for any $n$?

For any $k$ not in the form $6m+5$, is there $n$ such that $k$ is the smallest solution?

EDIT:

Let $f(n)$ be any polynomial $\mathbb N \rightarrow \mathbb N$. Let $\text{GCD}[f]$ denote the greatest common divisor of $\{f(n): n \in \mathbb N\}$. There is this long-standing conjecture by Bouniakowsky that $f$ is irreducible iff there are infinitely many numbers $n$ such that $\frac{f(n)}{\text{GCD}[f]}$ is prime.

If we suppose this is true, then the following statement implies a positive answer to question 3:

For any $k: k \not \equiv 5 \pmod 6$ there are natural numbers $\alpha,\beta$ such that $\text{GCD}[(\alpha n+\beta)^k+(\alpha n+\beta)-1]=1$, but $\text{GCD}[(\alpha n + \beta)^m+(\alpha n + \beta)-1] \neq 1$ for any $m<k: m \not \equiv 5 \pmod 6$.

Example: for $k=3$, one can let $\alpha = 15, \beta = 2$.

To my untrained eye, this seems approachable.

Title: Primes in the form $n^{k}+n-1$

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bounds: $1\le\text{n}\le10^3$ and $1\le\text{k}\le10$.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
\[Alpha] = 10^3;
\[Beta] = 10^1;
ParallelTable[
  If[PrimeQ[n^k + n - 1], {n, k}, Nothing], {n, 1, \[Alpha]}, {k,
   1, \[Beta]}] //. {} -> Nothing

Running the code gives:

Out[1]={{{2, 1}, {2, 2}, {2, 4}, {2, 8}}, {{3, 1}, {3, 2}, {3, 3}, {3,
   4}, {3, 8}, {3, 10}}, {{4, 1}, {4, 2}, {4, 3}, {4, 6}, {4, 8}, {4,
   9}}, {{5, 2}, {5, 6}, {5, 10}}, {{6, 1}, {6, 2}, {6, 4}, {6,
   7}, {6, 10}}, {{7, 1}, {7, 3}}, {{8, 2}, {8, 6}, {8, 10}}, {{9,
   1}, {9, 2}, {9, 4}, {9, 7}, {9, 10}}, {{10, 1}, {10, 2}, {10,
   3}, {10, 4}, {10, 9}}, {{11, 2}, {11, 8}}, {{12, 1}, {12,
   4}}, {{13, 2}, {13, 4}, {13, 10}}, {{15, 1}, {15, 2}, {15, 3}, {15,
    9}}, {{16, 1}, {16, 2}, {16, 3}, {16, 4}, {16, 8}, {16,
   10}}, {{17, 4}, {17, 8}}, {{18, 3}, {18, 6}}, {{19, 1}, {19,
   2}}, {{20, 2}, {20, 4}}, {{21, 1}, {21, 2}, {21, 3}}, {{22,
   1}, {22, 7}}, {{23, 4}, {23, 6}}, {{24, 1}, {24, 2}, {24,
   10}}, {{25, 3}, {25, 8}}, {{26, 2}, {26, 4}, {26, 10}}, {{27,
   1}, {27, 3}, {27, 8}}, {{28, 2}, {28, 4}}, {{29, 10}}, {{30,
   1}, {30, 2}, {30, 7}}, {{31, 1}, {31, 2}, {31, 4}, {31, 6}, {31,
   10}}, {{33, 3}, {33, 4}}, {{34, 1}, {34, 6}}, {{35, 2}}, {{36,
   1}, {36, 3}, {36, 6}}, {{37, 1}, {37, 7}}, {{38, 2}}, {{39,
   2}, {39, 3}}, {{40, 1}, {40, 4}, {40, 10}}, {{41, 2}}, {{42,
   1}}, {{43, 3}, {43, 4}}, {{44, 2}, {44, 4}}, {{45, 1}, {45,
   2}}, {{46, 2}, {46, 3}, {46, 8}}, {{48, 2}}, {{49, 1}, {49,
   7}, {49, 9}}, {{50, 2}, {50, 6}, {50, 8}}, {{51, 1}, {51, 3}, {51,
   6}, {51, 10}}, {{52, 1}, {52, 3}}, {{53, 2}, {53, 6}}, {{54,
   1}, {54, 2}, {54, 4}}, {{55, 1}, {55, 2}, {55, 3}}, {{56, 2}, {56,
   6}}, {{57, 1}, {57, 9}}, {{58, 4}}, {{59, 2}, {59, 10}}, {{60,
   2}, {60, 6}}, {{61, 7}, {61, 9}}, {{62, 8}}, {{63, 3}, {63,
   6}}, {{64, 1}, {64, 2}}, {{65, 2}}, {{66, 1}, {66, 2}, {66,
   8}, {66, 9}}, {{68, 2}, {68, 8}}, {{69, 1}, {69, 6}}, {{70,
   1}, {70, 2}, {70, 7}, {70, 9}}, {{72, 4}}, {{73, 3}, {73,
   6}}, {{74, 8}}, {{75, 1}, {75, 6}}, {{76, 1}, {76, 2}, {76,
   6}, {76, 10}}, {{77, 4}, {77, 8}}, {{78, 3}}, {{79, 1}, {79,
   9}}, {{81, 3}}, {{82, 1}}, {{83, 2}, {83, 6}}, {{84, 1}, {84,
   8}}, {{85, 2}}, {{86, 2}}, {{87, 1}, {87, 3}, {87, 8}}, {{88,
   8}}, {{89, 2}, {89, 10}}, {{90, 1}, {90, 10}}, {{91, 1}, {91,
   6}}, {{92, 4}}, {{93, 2}, {93, 3}, {93, 4}, {93, 7}}, {{94,
   2}, {94, 3}}, {{96, 1}, {96, 2}, {96, 3}, {96, 6}, {96, 10}}, {{97,
    1}}, {{98, 4}}, {{99, 1}, {99, 10}}, {{100, 1}, {100, 2}, {100,
   3}, {100, 7}, {100, 8}}, {{101, 2}}, {{103, 2}, {103, 3}, {103,
   10}}, {{105, 3}, {105, 4}, {105, 7}, {105, 9}}, {{106, 1}}, {{108,
   6}}, {{109, 3}}, {{110, 4}}, {{111, 6}, {111, 8}}, {{112, 1}, {112,
    3}, {112, 9}}, {{114, 1}, {114, 2}, {114, 9}}, {{115, 1}, {115,
   2}, {115, 3}, {115, 7}}, {{116, 6}}, {{117, 1}, {117, 3}, {117,
   4}, {117, 8}}, {{118, 6}, {118, 7}, {118, 8}}, {{119, 4}}, {{120,
   1}, {120, 2}, {120, 3}, {120, 4}, {120, 10}}, {{121, 1}, {121,
   9}}, {{122, 4}, {122, 8}}, {{124, 3}, {124, 10}}, {{125, 2}, {125,
   4}, {125, 8}}, {{126, 1}, {126, 2}, {126, 8}}, {{127, 3}}, {{129,
   1}, {129, 3}}, {{130, 2}}, {{131, 2}}, {{132, 1}, {132, 4}, {132,
   7}, {132, 8}}, {{133, 8}}, {{134, 2}, {134, 10}}, {{135, 1}, {135,
   3}, {135, 6}}, {{136, 1}}, {{138, 2}}, {{139, 1}, {139, 3}}, {{140,
    2}, {140, 8}}, {{141, 1}, {141, 2}}, {{142, 1}, {142, 7}, {142,
   9}}, {{143, 4}, {143, 6}}, {{144, 2}, {144, 9}}, {{145, 2}, {145,
   3}}, {{146, 8}}, {{147, 1}, {147, 9}}, {{148, 2}, {148, 6}}, {{149,
    2}, {149, 8}}, {{150, 3}}, {{151, 3}}, {{153, 2}, {153,
   8}}, {{154, 1}, {154, 2}}, {{155, 2}, {155, 10}}, {{156,
   1}}, {{157, 1}, {157, 4}}, {{158, 2}}, {{159, 1}, {159, 2}, {159,
   6}, {159, 7}}, {{160, 2}, {160, 8}}, {{162, 8}}, {{163, 2}, {163,
   7}}, {{164, 2}, {164, 4}, {164, 8}, {164, 10}}, {{165, 3}, {165,
   8}, {165, 9}}, {{166, 1}, {166, 3}, {166, 9}}, {{168, 8}, {168,
   10}}, {{169, 1}, {169, 2}, {169, 7}}, {{171, 2}, {171, 3}, {171,
   8}}, {{172, 9}}, {{174, 1}, {174, 2}, {174, 9}}, {{175, 1}}, {{176,
    2}}, {{177, 1}}, {{178, 3}, {178, 10}}, {{180, 1}, {180, 2}, {180,
    7}, {180, 8}}, {{181, 2}, {181, 8}}, {{182, 4}}, {{183, 6}, {183,
   8}}, {{184, 1}, {184, 2}, {184, 7}}, {{185, 10}}, {{186,
   2}}, {{187, 1}}, {{188, 2}, {188, 6}, {188, 8}}, {{189, 3}, {189,
   6}}, {{190, 1}}, {{191, 2}, {191, 6}}, {{192, 1}, {192, 3}}, {{193,
    2}}, {{195, 1}, {195, 2}}, {{196, 2}, {196, 7}}, {{198,
   3}}, {{199, 1}, {199, 2}, {199, 3}, {199, 9}}, {{201, 1}, {201,
   4}, {201, 6}}, {{202, 3}}, {{203, 2}}, {{204, 3}, {204, 4}}, {{205,
    1}, {205, 4}}, {{206, 2}, {206, 6}}, {{208, 7}}, {{209, 2}, {209,
   8}}, {{210, 1}}, {{211, 1}, {211, 7}}, {{215, 2}}, {{216, 1}, {216,
    3}, {216, 8}, {216, 9}}, {{217, 1}}, {{218, 2}}, {{219,
   2}}, {{220, 1}, {220, 2}}, {{221, 8}}, {{222, 1}}, {{223,
   3}}, {{225, 1}, {225, 2}}, {{228, 3}}, {{229, 1}, {229, 4}, {229,
   9}}, {{230, 2}, {230, 6}, {230, 8}, {230, 10}}, {{231, 1}, {231,
   2}, {231, 4}, {231, 6}}, {{232, 1}}, {{233, 2}}, {{234, 1}, {234,
   6}}, {{235, 8}}, {{236, 2}}, {{240, 1}, {240, 2}}, {{241, 2}, {241,
    3}}, {{243, 3}}, {{244, 1}, {244, 2}, {244, 9}}, {{246, 1}, {246,
   2}, {246, 3}}, {{247, 7}}, {{248, 2}, {248, 6}, {248, 8}}, {{250,
   1}}, {{252, 1}}, {{253, 8}}, {{254, 6}, {254, 8}}, {{255, 1}, {255,
    6}, {255, 7}}, {{256, 9}}, {{258, 2}, {258, 3}}, {{259, 2}, {259,
   3}, {259, 7}}, {{260, 6}, {260, 8}}, {{261, 1}}, {{262, 1}}, {{263,
    2}}, {{264, 2}}, {{265, 2}, {265, 8}}, {{266, 4}}, {{268,
   2}}, {{271, 1}, {271, 6}}, {{274, 1}}, {{275, 4}}, {{276,
   7}}, {{279, 1}, {279, 4}}, {{280, 8}}, {{281, 2}}, {{282,
   1}}, {{283, 3}}, {{285, 1}, {285, 2}, {285, 3}}, {{286, 1}, {286,
   3}, {286, 4}}, {{288, 2}, {288, 4}}, {{289, 1}, {289, 6}}, {{290,
   2}, {290, 4}, {290, 8}}, {{291, 9}}, {{292, 4}, {292, 8}}, {{293,
   4}}, {{294, 1}, {294, 2}}, {{296, 2}}, {{297, 1}}, {{298,
   2}}, {{300, 1}, {300, 3}}, {{301, 1}, {301, 2}}, {{303, 2}}, {{304,
    1}, {304, 4}}, {{305, 2}, {305, 6}}, {{306, 2}}, {{307, 1}, {307,
   7}}, {{308, 10}}, {{309, 1}, {309, 2}, {309, 4}}, {{310, 1}, {310,
   3}, {310, 8}}, {{312, 8}}, {{313, 3}}, {{314, 2}}, {{316, 1}, {316,
    3}}, {{318, 4}}, {{319, 2}, {319, 3}}, {{321, 1}, {321,
   4}}, {{322, 1}, {322, 4}}, {{323, 2}, {323, 6}}, {{324, 1}, {324,
   10}}, {{326, 8}}, {{327, 1}, {327, 7}, {327, 8}}, {{328,
   6}}, {{329, 4}}, {{330, 1}, {330, 2}, {330, 4}, {330, 9}}, {{331,
   1}, {331, 2}, {331, 6}}, {{334, 3}, {334, 4}}, {{335, 2}}, {{336,
   10}}, {{337, 1}}, {{339, 1}, {339, 2}, {339, 10}}, {{342,
   1}}, {{343, 2}, {343, 3}}, {{344, 4}}, {{346, 1}, {346, 9}}, {{348,
    3}, {348, 6}}, {{349, 2}, {349, 6}}, {{350, 2}, {350, 4}}, {{351,
   1}, {351, 2}, {351, 4}, {351, 10}}, {{352, 3}, {352, 7}}, {{353,
   4}}, {{354, 2}}, {{355, 1}}, {{356, 8}}, {{358, 2}}, {{360,
   1}, {360, 2}, {360, 3}, {360, 8}}, {{361, 2}}, {{363, 4}}, {{364,
   1}, {364, 2}}, {{366, 3}, {366, 4}, {366, 10}}, {{367, 1}, {367,
   3}, {367, 8}}, {{370, 1}, {370, 9}}, {{371, 6}}, {{372, 1}}, {{373,
    2}, {373, 4}, {373, 6}}, {{374, 2}}, {{375, 6}}, {{376, 1}, {376,
   3}}, {{378, 2}}, {{379, 1}, {379, 3}}, {{380, 2}, {380, 6}}, {{381,
    1}, {381, 3}, {381, 7}}, {{385, 1}, {385, 2}, {385, 6}, {385,
   7}, {385, 8}}, {{386, 2}, {386, 6}}, {{387, 1}, {387, 7}, {387,
   9}}, {{390, 4}}, {{391, 2}, {391, 3}}, {{393, 2}, {393, 6}, {393,
   8}}, {{394, 1}}, {{395, 2}, {395, 10}}, {{396, 2}}, {{397,
   4}}, {{398, 4}, {398, 8}}, {{399, 1}, {399, 4}}, {{401, 2}, {401,
   6}}, {{405, 1}, {405, 2}}, {{406, 1}, {406, 4}, {406, 7}}, {{407,
   4}}, {{408, 2}, {408, 4}, {408, 7}}, {{409, 4}, {409, 8}}, {{411,
   1}}, {{412, 1}, {412, 3}}, {{414, 1}, {414, 6}}, {{415, 1}}, {{417,
    3}}, {{418, 2}, {418, 3}}, {{419, 2}}, {{420, 1}, {420, 2}, {420,
   6}}, {{421, 3}}, {{423, 2}, {423, 8}}, {{424, 3}, {424, 8}}, {{426,
    3}, {426, 8}}, {{427, 1}}, {{428, 2}}, {{429, 1}, {429, 3}, {429,
   8}}, {{430, 1}, {430, 3}}, {{431, 2}}, {{432, 1}, {432, 3}}, {{433,
    2}}, {{434, 8}}, {{435, 9}}, {{437, 8}}, {{438, 6}}, {{439,
   1}, {439, 10}}, {{440, 8}}, {{441, 1}}, {{442, 1}}, {{443,
   6}}, {{444, 1}, {444, 3}}, {{445, 2}}, {{446, 8}}, {{447,
   8}}, {{448, 2}}, {{449, 2}}, {{450, 2}, {450, 3}}, {{451,
   8}}, {{453, 2}}, {{454, 1}, {454, 3}}, {{455, 2}, {455, 4}}, {{456,
    1}, {456, 2}, {456, 7}}, {{459, 3}}, {{460, 1}, {460, 4}, {460,
   6}}, {{461, 2}}, {{462, 7}}, {{463, 2}, {463, 4}}, {{465,
   1}}, {{466, 3}, {466, 6}}, {{468, 2}, {468, 3}}, {{469, 1}, {469,
   3}, {469, 4}, {469, 6}}, {{471, 1}, {471, 2}, {471, 6}}, {{472,
   3}, {472, 4}, {472, 9}}, {{473, 2}, {473, 8}}, {{474, 1}, {474,
   2}}, {{475, 2}}, {{477, 1}}, {{478, 2}}, {{481, 2}, {481,
   3}}, {{482, 4}}, {{483, 4}, {483, 7}}, {{484, 1}, {484, 3}}, {{486,
    1}, {486, 2}, {486, 3}, {486, 9}}, {{489, 1}, {489, 3}, {489,
   9}, {489, 10}}, {{490, 2}}, {{492, 1}}, {{493, 6}}, {{494,
   2}, {494, 6}}, {{495, 2}, {495, 9}}, {{496, 1}, {496, 2}}, {{499,
   1}, {499, 2}, {499, 7}}, {{500, 2}}, {{501, 2}}, {{502, 4}}, {{504,
    2}}, {{505, 1}, {505, 3}}, {{506, 2}, {506, 4}}, {{507, 1}, {507,
   4}}, {{508, 6}}, {{510, 1}, {510, 2}, {510, 3}}, {{511, 1}, {511,
   2}}, {{513, 4}}, {{514, 3}, {514, 10}}, {{515, 2}, {515,
   6}}, {{516, 1}, {516, 4}}, {{517, 1}}, {{518, 2}}, {{519, 2}, {519,
    8}}, {{520, 1}, {520, 4}, {520, 6}}, {{521, 4}, {521, 10}}, {{522,
    3}}, {{524, 4}}, {{525, 1}, {525, 6}}, {{526, 1}}, {{528,
   2}, {528, 3}, {528, 7}, {528, 10}}, {{529, 9}}, {{530, 2}}, {{531,
   1}}, {{532, 1}, {532, 3}, {532, 7}}, {{534, 4}}, {{535, 1}, {535,
   3}, {535, 4}}, {{537, 9}}, {{538, 3}, {538, 4}, {538, 7}}, {{539,
   4}}, {{540, 6}, {540, 7}}, {{541, 2}, {541, 3}}, {{542, 4}}, {{543,
    8}}, {{544, 1}, {544, 2}, {544, 6}, {544, 8}, {544, 9}, {544,
   10}}, {{546, 1}, {546, 7}}, {{547, 1}}, {{548, 2}}, {{549,
   1}, {549, 2}}, {{550, 2}}, {{551, 2}, {551, 6}}, {{552, 1}}, {{553,
    3}, {553, 4}}, {{554, 4}, {554, 8}}, {{555, 1}, {555, 4}, {555,
   6}, {555, 8}}, {{556, 4}}, {{558, 3}, {558, 10}}, {{559, 1}, {559,
   6}, {559, 9}}, {{560, 2}}, {{561, 2}, {561, 4}}, {{562, 1}, {562,
   3}, {562, 4}}, {{564, 9}}, {{565, 1}, {565, 8}}, {{566,
   10}}, {{567, 4}, {567, 8}}, {{568, 3}}, {{569, 2}}, {{571,
   2}}, {{572, 4}}, {{573, 2}, {573, 4}}, {{574, 3}, {574, 4}}, {{576,
    1}}, {{577, 1}, {577, 4}}, {{578, 2}}, {{579, 3}}, {{580,
   8}, {580, 10}}, {{581, 2}, {581, 4}}, {{582, 1}, {582, 3}}, {{585,
   6}}, {{586, 1}, {586, 9}, {586, 10}}, {{588, 2}}, {{589,
   2}}, {{590, 4}}, {{591, 1}, {591, 2}}, {{593, 10}}, {{594,
   1}, {594, 2}, {594, 7}, {594, 10}}, {{595, 2}, {595, 8}}, {{596,
   2}}, {{597, 1}}, {{598, 2}, {598, 4}}, {{599, 8}}, {{601, 1}, {601,
    10}}, {{604, 2}}, {{605, 4}}, {{606, 2}}, {{607, 1}}, {{608,
   4}, {608, 10}}, {{609, 1}, {609, 10}}, {{610, 2}, {610, 3}}, {{612,
    1}, {612, 4}, {612, 9}}, {{613, 8}}, {{615, 1}, {615, 3}, {615,
   8}}, {{616, 1}, {616, 2}, {616, 10}}, {{618, 2}}, {{619, 1}, {619,
   4}}, {{623, 4}, {623, 6}}, {{624, 2}, {624, 9}}, {{625, 1}, {625,
   2}, {625, 4}, {625, 9}, {625, 10}}, {{628, 6}}, {{629, 2}, {629,
   4}, {629, 10}}, {{630, 1}, {630, 8}, {630, 9}}, {{631, 3}, {631,
   10}}, {{633, 2}}, {{634, 6}}, {{635, 10}}, {{637, 8}}, {{639,
   1}, {639, 2}, {639, 4}}, {{640, 1}, {640, 2}}, {{641, 8}}, {{642,
   1}, {642, 3}}, {{643, 8}}, {{644, 2}, {644, 4}}, {{645, 1}, {645,
   7}}, {{646, 1}, {646, 2}}, {{648, 2}}, {{649, 1}, {649, 7}, {649,
   10}}, {{651, 1}, {651, 2}, {651, 8}}, {{652, 1}, {652, 4}}, {{654,
   1}, {654, 2}}, {{655, 2}}, {{656, 6}}, {{658, 7}}, {{659, 2}, {659,
    6}, {659, 8}}, {{660, 1}, {660, 3}, {660, 6}}, {{661, 1}, {661,
   3}}, {{663, 6}}, {{664, 1}, {664, 4}}, {{665, 10}}, {{666,
   3}, {666, 4}, {666, 7}}, {{667, 3}}, {{668, 2}, {668, 6}}, {{670,
   2}, {670, 6}, {670, 9}}, {{673, 2}}, {{674, 10}}, {{675,
   8}}, {{676, 2}, {676, 8}, {676, 9}}, {{681, 1}, {681, 10}}, {{683,
   2}}, {{684, 1}, {684, 3}, {684, 4}, {684, 7}}, {{685, 3}}, {{686,
   2}, {686, 10}}, {{687, 1}, {687, 3}}, {{688, 3}, {688, 4}, {688,
   6}, {688, 10}}, {{690, 3}, {690, 8}}, {{691, 1}, {691, 2}}, {{693,
   2}, {693, 10}}, {{694, 3}}, {{695, 2}}, {{697, 3}}, {{699,
   2}}, {{700, 1}}, {{703, 6}}, {{705, 1}, {705, 2}}, {{706, 2}, {706,
    8}}, {{708, 2}}, {{709, 2}}, {{710, 10}}, {{711, 9}}, {{712,
   1}}, {{714, 1}, {714, 4}, {714, 8}}, {{715, 1}, {715, 2}, {715,
   7}}, {{716, 2}}, {{717, 1}, {717, 4}}, {{720, 1}, {720, 2}}, {{721,
    4}, {721, 8}}, {{722, 4}}, {{723, 3}}, {{724, 1}, {724,
   2}}, {{726, 1}, {726, 3}}, {{727, 1}, {727, 4}}, {{728, 2}, {728,
   10}}, {{730, 1}, {730, 10}}, {{731, 4}, {731, 10}}, {{733,
   3}}, {{736, 1}}, {{738, 10}}, {{739, 2}, {739, 3}, {739,
   9}}, {{741, 1}, {741, 4}, {741, 7}}, {{742, 1}, {742, 3}}, {{743,
   2}, {743, 4}}, {{744, 1}}, {{745, 1}}, {{746, 2}}, {{747,
   1}}, {{748, 4}}, {{750, 1}, {750, 2}, {750, 3}}, {{751, 3}, {751,
   8}}, {{753, 2}, {753, 3}}, {{754, 2}}, {{755, 10}}, {{756,
   1}}, {{757, 7}}, {{759, 7}}, {{761, 2}}, {{762, 1}, {762,
   4}}, {{763, 2}, {763, 3}, {763, 4}, {763, 8}}, {{765, 2}}, {{766,
   1}, {766, 6}}, {{768, 10}}, {{769, 2}, {769, 3}, {769, 9}}, {{770,
   6}}, {{771, 6}}, {{772, 1}}, {{774, 6}, {774, 8}}, {{775, 1}, {775,
    6}, {775, 10}}, {{776, 2}, {776, 4}}, {{777, 1}, {777, 3}}, {{778,
    8}}, {{779, 2}, {779, 6}}, {{780, 1}, {780, 2}, {780, 4}}, {{781,
   2}, {781, 8}}, {{783, 3}}, {{784, 1}, {784, 3}}, {{786, 1}, {786,
   2}, {786, 4}}, {{787, 7}}, {{789, 9}}, {{790, 1}, {790, 6}, {790,
   9}}, {{792, 1}, {792, 8}}, {{794, 2}}, {{795, 8}}, {{796, 6}, {796,
    8}}, {{797, 8}}, {{798, 2}, {798, 6}, {798, 7}}, {{799,
   1}}, {{800, 6}, {800, 8}}, {{801, 1}}, {{803, 2}, {803, 4}, {803,
   10}}, {{804, 1}, {804, 2}, {804, 7}}, {{805, 1}}, {{807, 1}, {807,
   4}, {807, 8}}, {{808, 8}}, {{809, 2}}, {{810, 1}, {810, 9}}, {{811,
    1}}, {{813, 6}}, {{814, 1}, {814, 2}, {814, 3}, {814, 4}, {814,
   9}}, {{815, 2}, {815, 8}, {815, 10}}, {{816, 2}, {816, 3}, {816,
   9}}, {{819, 1}, {819, 3}}, {{825, 2}, {825, 10}}, {{828,
   3}}, {{829, 1}, {829, 3}}, {{831, 6}}, {{832, 1}, {832, 3}, {832,
   4}}, {{833, 2}, {833, 8}}, {{834, 1}, {834, 4}, {834, 9}}, {{835,
   1}, {835, 3}}, {{836, 10}}, {{837, 3}}, {{838, 2}}, {{839,
   8}}, {{841, 3}}, {{843, 3}}, {{845, 2}, {845, 4}}, {{846,
   10}}, {{847, 1}}, {{848, 2}, {848, 6}}, {{849, 1}}, {{850,
   1}, {850, 4}, {850, 9}}, {{852, 8}}, {{853, 10}}, {{855,
   1}}, {{856, 2}, {856, 10}}, {{857, 4}}, {{860, 4}}, {{861,
   1}, {861, 3}, {861, 4}}, {{862, 1}}, {{863, 2}}, {{866, 6}}, {{867,
    1}}, {{868, 6}}, {{869, 6}}, {{870, 7}}, {{871, 1}, {871,
   3}, {871, 8}}, {{873, 2}, {873, 4}}, {{874, 1}}, {{875, 6}, {875,
   8}}, {{876, 4}}, {{877, 1}, {877, 7}}, {{879, 3}}, {{880, 1}, {880,
    6}, {880, 10}}, {{881, 2}}, {{882, 3}}, {{883, 4}}, {{884,
   2}}, {{885, 2}}, {{888, 3}}, {{889, 1}, {889, 2}}, {{890, 2}, {890,
    4}, {890, 6}}, {{891, 4}}, {{892, 1}}, {{894, 1}, {894, 6}, {894,
   10}}, {{895, 1}}, {{897, 3}}, {{898, 10}}, {{901, 1}, {901,
   2}, {901, 8}}, {{906, 1}, {906, 2}, {906, 8}}, {{909, 3}, {909,
   8}, {909, 9}}, {{910, 3}}, {{912, 1}}, {{913, 4}, {913, 7}}, {{915,
    2}}, {{916, 1}, {916, 8}}, {{917, 8}}, {{918, 6}}, {{919,
   8}, {919, 9}}, {{921, 2}, {921, 8}}, {{922, 4}}, {{923, 2}}, {{924,
    1}, {924, 4}, {924, 7}}, {{925, 2}, {925, 4}}, {{926, 6}}, {{928,
   6}}, {{930, 2}, {930, 9}}, {{931, 1}, {931, 6}}, {{934, 1}}, {{936,
    1}}, {{937, 1}, {937, 7}}, {{938, 4}, {938, 10}}, {{939, 1}, {939,
    2}, {939, 8}}, {{940, 1}, {940, 3}, {940, 9}, {940, 10}}, {{941,
   2}}, {{943, 3}}, {{944, 2}}, {{945, 1}, {945, 9}}, {{946, 2}, {946,
    4}}, {{948, 3}, {948, 4}}, {{950, 2}, {950, 6}, {950, 8}}, {{951,
   1}, {951, 4}, {951, 8}}, {{952, 4}}, {{953, 4}}, {{954, 1}}, {{955,
    2}, {955, 4}, {955, 7}}, {{956, 2}}, {{957, 1}}, {{960,
   6}}, {{961, 8}}, {{963, 2}}, {{965, 2}}, {{966, 1}, {966,
   2}}, {{967, 1}}, {{968, 2}}, {{969, 3}, {969, 8}}, {{970, 3}, {970,
    6}, {970, 9}}, {{972, 3}}, {{974, 2}}, {{975, 1}}, {{976,
   1}, {976, 2}, {976, 6}}, {{979, 4}, {979, 8}, {979, 9}}, {{980,
   4}}, {{981, 2}, {981, 4}}, {{982, 3}, {982, 8}}, {{984, 2}, {984,
   3}, {984, 6}}, {{987, 1}, {987, 3}}, {{988, 6}}, {{989, 2}, {989,
   10}}, {{990, 1}, {990, 3}}, {{991, 3}}, {{992, 4}}, {{994,
   1}, {994, 2}, {994, 6}}, {{996, 2}, {996, 3}, {996, 4}, {996,
   6}}, {{997, 1}}, {{998, 2}}, {{999, 1}, {999, 9}}, {{1000,
   1}, {1000, 2}}}

So, we can see that using the bounds $1\le\text{n}\le10^3$ and $1\le\text{k}\le10$ we found $1419$ solutions. That number is found by using the following Mathematica-code:

In[2]:=Clear["Global`*"];
\[Alpha] = 10^3;
\[Beta] = 10^1;
f = Total@*Map[Length];
f[ParallelTable[
   If[PrimeQ[n^k + n - 1], {n, k}, Nothing], {n, 1, \[Alpha]}, {k,
    1, \[Beta]}] //. {} -> Nothing]

Out[2]=1419

Extending the bounds to $1\le\text{n}\le10^4$ and $1\le\text{k}\le10^2$ gives $19235$ solutions.

I did a search to find solutions for $\text{n}=107$. I did find a solution for $1\le\text{k}\le5\cdot10^3$ because when $\text{k}=1400$ the number $\text{n}^\text{k}+\text{n}-1$ is prime.

I also searched for solutions of $\text{n}$ and $\text{k}$ such that $\text{k}\ne6\text{m}+5$ (where $\text{m}\in\mathbb{N}$). I did find solutions for $1\le\text{n}\le10^2$ and $1\le\text{k}\le10$:

In[3]:=Clear["Global`*"];
\[Alpha] = 10^2;
\[Beta] = 10^1;
ParallelTable[
  If[TrueQ[PrimeQ[n^k + n - 1] && IntegerQ[(k - 5)/6] == False], {n,
    k}, Nothing], {n, 1, \[Alpha]}, {k, 1, \[Beta]}] //. {} -> Nothing

Out[3]={{{2, 1}, {2, 2}, {2, 4}, {2, 8}}, {{3, 1}, {3, 2}, {3, 3}, {3,
   4}, {3, 8}, {3, 10}}, {{4, 1}, {4, 2}, {4, 3}, {4, 6}, {4, 8}, {4,
   9}}, {{5, 2}, {5, 6}, {5, 10}}, {{6, 1}, {6, 2}, {6, 4}, {6,
   7}, {6, 10}}, {{7, 1}, {7, 3}}, {{8, 2}, {8, 6}, {8, 10}}, {{9,
   1}, {9, 2}, {9, 4}, {9, 7}, {9, 10}}, {{10, 1}, {10, 2}, {10,
   3}, {10, 4}, {10, 9}}, {{11, 2}, {11, 8}}, {{12, 1}, {12,
   4}}, {{13, 2}, {13, 4}, {13, 10}}, {{15, 1}, {15, 2}, {15, 3}, {15,
    9}}, {{16, 1}, {16, 2}, {16, 3}, {16, 4}, {16, 8}, {16,
   10}}, {{17, 4}, {17, 8}}, {{18, 3}, {18, 6}}, {{19, 1}, {19,
   2}}, {{20, 2}, {20, 4}}, {{21, 1}, {21, 2}, {21, 3}}, {{22,
   1}, {22, 7}}, {{23, 4}, {23, 6}}, {{24, 1}, {24, 2}, {24,
   10}}, {{25, 3}, {25, 8}}, {{26, 2}, {26, 4}, {26, 10}}, {{27,
   1}, {27, 3}, {27, 8}}, {{28, 2}, {28, 4}}, {{29, 10}}, {{30,
   1}, {30, 2}, {30, 7}}, {{31, 1}, {31, 2}, {31, 4}, {31, 6}, {31,
   10}}, {{33, 3}, {33, 4}}, {{34, 1}, {34, 6}}, {{35, 2}}, {{36,
   1}, {36, 3}, {36, 6}}, {{37, 1}, {37, 7}}, {{38, 2}}, {{39,
   2}, {39, 3}}, {{40, 1}, {40, 4}, {40, 10}}, {{41, 2}}, {{42,
   1}}, {{43, 3}, {43, 4}}, {{44, 2}, {44, 4}}, {{45, 1}, {45,
   2}}, {{46, 2}, {46, 3}, {46, 8}}, {{48, 2}}, {{49, 1}, {49,
   7}, {49, 9}}, {{50, 2}, {50, 6}, {50, 8}}, {{51, 1}, {51, 3}, {51,
   6}, {51, 10}}, {{52, 1}, {52, 3}}, {{53, 2}, {53, 6}}, {{54,
   1}, {54, 2}, {54, 4}}, {{55, 1}, {55, 2}, {55, 3}}, {{56, 2}, {56,
   6}}, {{57, 1}, {57, 9}}, {{58, 4}}, {{59, 2}, {59, 10}}, {{60,
   2}, {60, 6}}, {{61, 7}, {61, 9}}, {{62, 8}}, {{63, 3}, {63,
   6}}, {{64, 1}, {64, 2}}, {{65, 2}}, {{66, 1}, {66, 2}, {66,
   8}, {66, 9}}, {{68, 2}, {68, 8}}, {{69, 1}, {69, 6}}, {{70,
   1}, {70, 2}, {70, 7}, {70, 9}}, {{72, 4}}, {{73, 3}, {73,
   6}}, {{74, 8}}, {{75, 1}, {75, 6}}, {{76, 1}, {76, 2}, {76,
   6}, {76, 10}}, {{77, 4}, {77, 8}}, {{78, 3}}, {{79, 1}, {79,
   9}}, {{81, 3}}, {{82, 1}}, {{83, 2}, {83, 6}}, {{84, 1}, {84,
   8}}, {{85, 2}}, {{86, 2}}, {{87, 1}, {87, 3}, {87, 8}}, {{88,
   8}}, {{89, 2}, {89, 10}}, {{90, 1}, {90, 10}}, {{91, 1}, {91,
   6}}, {{92, 4}}, {{93, 2}, {93, 3}, {93, 4}, {93, 7}}, {{94,
   2}, {94, 3}}, {{96, 1}, {96, 2}, {96, 3}, {96, 6}, {96, 10}}, {{97,
    1}}, {{98, 4}}, {{99, 1}, {99, 10}}, {{100, 1}, {100, 2}, {100,
   3}, {100, 7}, {100, 8}}}


问:
Alternate method for finding ways of distribution

Find the number of ways in which $n$ distinct balls can be distributed between $n$ children such that exactly one child doesn't get a ball.

This can be solved easily by looking at it as division and distribution.

Here, $n-2$ children get exactly $1$ ball, one child gets $2$ balls and one child gets no ball. The number of possible divisions is just

$$\frac{n!}{0!2!(1!)^{n-2}(n-2)!}=\frac{n!}{(n-2)!2!}$$

Hence the number of possible distributions is

$$\frac{n!}{(n-2)!2!}\times n!=^nC_2\times n!$$

This answer looks suspiciously close to the answers we would get by using the normal selection and permutation method. However I couldn't figure out any "normal" way. Can you help me out?

答:
Make $n$ piles with $0,2, 1,1,1...$  balls with the pile of $2$ chosen in $\binom n 2$ ways, and the piles permuted in $n!$  ways giving $\binom n 2 n!$ arrangements

Line up the kids in (say) alphabetical order and distribute.


问:
Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $
As the title described, I was trying to find an alternative proof to

If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.

Here's the proof that I've found (I'm sorry, I forgot where I got it):

Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.

Now since I love to punish myself, I tried to find a harder proof as such:

We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2)  +2t} + 1
$$ which simplifies to  $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$

Now how do I prove the sextic polynomial inequality above (which is true)?

答:
Alternative proof  via C-S and AM-GM.

$$x^3+y^3\ge \sqrt 2 xy \iff \frac{x^2}{y} + \frac{y^2}{x} \ge \sqrt 2$$

$$1 = (x^2 + y^2)^2 \le \left( \frac{x^2}{y} + \frac{y^2}{x} \right) \cdot (x^2 y + xy^2) \tag {CS}$$

$$= \left( \frac{x^2}{y} + \frac{y^2}{x} \right) \sqrt{x^2y^2(x+y)^2}\\
= \left( \frac{x^2}{y} + \frac{y^2}{x} \right) \sqrt{(xy)^2(1+2xy)}\\
\le \left( \frac{x^2}{y} + \frac{y^2}{x} \right) \sqrt{\frac 14 \left(1+2\frac 12\right)}\\$$

Done.


问:
Check if $u_y+u^2 u_x = 0$ with rectangular initial data has a shock

I have the p.d.e. $$u_y+u^2 u_x =0$$ $x \in \mathbb{R}$, $t>0$
with $$u(x,0)=h(x)= \begin{cases} 0 & x<0 \\ 1 & 0<x<1 \\ 0 & x>1 \end{cases}$$
Check if a shock forms and solve the p.d.e.

My attempt, the characteristic curves are $(s,0,h(s))$, so I have $$\frac{dy}{dt}=1 \implies y=t$$
$$\frac{du}{dt}=0 \implies u=h(s)$$
$$\frac{dx}{dt}=u^2 \implies x=h^2(s)t+s$$

So the solution is $$u=h(x-u^2y)$$ $$u= \begin{cases} 0 & x<0 \\ 1 & t <x <t+1 \\ 0 & x>1 \end{cases}$$
Now this is where I am stuck, how do I check if a shock forms?
And how do I proceed to solve the problem?

答:
To recognize the formation of singularities (shocks) in your equation you may look for the intersection of characteristics curves. It is known that any smooth solution of your problem must be constant along characteristic curves, and if there is such an intersection, singularities arise.

Now, from your characteristic system, it can be seen that
$$ \frac{dx}{dy} = u^2, $$
due to the constancy of $u$ along characteristics, this means that the projection of the charactericts curves are straight lines with slope given by $1/u^2(x,0)$. Try to draw them out and use your initial condition to see if those lines intersect at some point.

On the other hand, to find a solution for your problem, you should bear in mind something which is known in the literature as the "jump condition" or "Rankine-Hugoniot condition". You can find information about it (together with nice examples) in Joel Smoller book : "Shock waves and reaction diffusion equations" Chapter 15. This condition allows you to give discontinuous solutions for your problem whenever an specific relation between the speed of propagation of the discontinuity and the values of the solution at both sides of the discontinuity is satisfied.

Roughly speaking, you must ensure that if $s$ is the speed of propagation of the discontinuity, then
$$ s[u]=[u^2],$$
where $[u]= u_r - u_l$, $[u^2] = u_r^2 - u_l^2  $ and $u_r$, $u_l$ are the values of $u$ at the right and left of the discontinuity respectively.


问:
Enumerable topology
Let $X$ be a non-empty, fixed set $x_{0} \in X$ arbitrary. Show that $B = \{\{x\}: x \in X\backslash\{x_0\}\} \cup \{A\subseteq X: x_0\in A$ e $X\backslash A$ is enumerable $\}$ is a base for a topology over $X$.

Be $X$ a set and $B$ a collection of subsets of $X$ satisfying the following properties:

(1) For each $U_1$, $U_2$ and each $x\in U_1\cap U_2$, there is $U\in B$ such that $x \in U\subseteq U_1\cap U_2$.

(2) For each $x\in X$, there is $U\in B$ such as $x\in U$.

So the $B^{\ast}$, formed by the $B$ subcollection unions, is a topology about $X$ and $B$ is a basis for the $(X,B^{\ast})$topological space.

How can I prove that given $U_1$ and $U_2$ and a point at the intersection then there is $U$ at the intersection, such as $x$ is at $U$ ? My questions are whether I $U_1$ and $U_2$ arbitrary?

Also, how can I show (2), which always exists $x\in X$, is there $U\in B$ such that $x\in U$?

答:
You can prove something stronger, that is $U_1\cap U_2\in B\cup\{\varnothing\}$. Take $U_i\in B$, $i=1,2$, this means that either $U_i=\{x\}$ or $X\setminus U_1$ is enumerable, this gives you three cases:

$U_1=\{x_1\}, U_2=\{x_2\}$, this means that either $U_1\cap U_2=\varnothing$, or $x_1=x_2$ wich means $U_1=U_2=U_1\cap U_2$.

$U_1=\{x\}$, $X\setminus U_2$ is enumerable, then $U_1\cap U_2=\varnothing$ if $x\notin U_2$ or $U_1\cap U_2=U_1$ if $x\in U_2$.

$X\setminus U_1$ and $X\setminus U_2$ are enumerable, then $X\setminus (U_1\cap U_2)=(X\setminus U_1)\cup (X\setminus U_2)$, since the union of two enumerable sets is enumerable you are finished.

Title: Enumerable topology


问:
Query regarding a proof of translation invariance of lebesgue measue
Here $\mathcal{B}(\mathbb{R})$ will denote the borel sigma field in $\mathbb{R}$ and $\lambda$ denotes the legesbue measure.

I have already proved that $A+x\in\mathcal{B}(\mathbb{R})\ \forall A\in\mathcal{B}(\mathbb{R}),\ x\in\mathbb{R}$ by showing that the collection of "good sets" here $\{A:\ A+x\in\mathcal{B}(\mathbb{R})\}$ is a $\sigma$-field.

Now again I use good set trick and construct $S=\{A\in\mathcal{B}(\mathbb{R}):\ \lambda(A+x)=\lambda(A)\}$. It's easy to see that $S$ contains all the intervals in $\mathbb{R}$. I want to show that $S$ is a $\sigma$-field.

But I'm stuck with only one step i.e. if $A\in S$ then $A^c\in S$.

So we have $\lambda(A+x)=\lambda(A)$, we have to show that $\lambda(A^c+x)=\lambda((A+x)^c)=\lambda(A^c)$.

Here $\lambda$ is not finite measure, so I'm not getting any wayout to prove the above claim.

Can anyone help me in this regard? Thanks for assistance in advance.

答:
To prove the translation invariance of Lebesgue measure, the last step is a little more subtle, exactly to take care of sets of infinite measure. The  $\sigma$-field $S$ must be defined in a slightly different way. Let us see it.

Define
$$S=\{A\in\mathcal{B}(\mathbb{R}): \textrm{ for all } n \in \Bbb N,  \lambda((A\cap [-n,n])+x)=\lambda(A\cap [-n,n])\}$$
It easy to see that $S$ contains all the intervals in $\mathbb{R}$ and that $S$ is a $\sigma$-field. So $\mathcal{B}(\mathbb{R}) \subseteq S$ (in fact, $\mathcal{B}(\mathbb{R}) = S$).

So we have that, for all $A \in \mathcal{B}(\mathbb{R})$ and all $n \in \Bbb N$, $\lambda((A\cap [-n,n])+x)=\lambda(A\cap [-n,n])\}$.

Now, note that, for all $A \in \mathcal{B}(\mathbb{R})$, $A\cap [-n,n] \nearrow A$ and $(A\cap [-n,n])+x \nearrow A+x$. So we have
$$\lambda(A+x)= \lim_{n\to \infty}\lambda((A\cap [-n,n])+x)=\lim_{n\to \infty}\lambda(A\cap [-n,n])= \lambda(A) $$

Remark: Let us prove in detail that $S$ is closed under complement.
Suppose $A \in S$. It means that $A \in \mathcal{B}$ and, for each $n \in \Bbb N$, $\lambda((A\cap [-n,n])+x)=\lambda(A\cap [-n,n])\}$.

So we have that $A^c\in \mathcal{B}$ and for each $n \in \Bbb N$,
\begin{align*}
\lambda( (A^c\cap [-n,n]) +x) & = \lambda((A^c+x)\cap [-n+x,n+x]) \:\:\:\:\text{ by 1}\\
& = \lambda((A+x)^c\cap [-n+x,n+x]) \:\:\:\:\text{ by 2}\\
&= \lambda([-n+x,n+x]) - \lambda((A+x)\cap [-n+x,n+x]) \\
&= 2n - \lambda((A+x)\cap [-n+x,n+x]) \\
& = 2n - \lambda((A\cap [-n,n])+x)  \:\:\:\:\text{ by 3}\\
&= 2n -  \lambda(A\cap [-n,n]) \\
&= \lambda([-n,n]) -  \lambda(A\cap [-n,n]) \\
& = \lambda(A^c\cap [-n,n])
\end{align*}
where we used:

$(A^c\cap [-n,n]) +x = (A^c+x)\cap [-n+x,n+x]$

$A^c+x =(A+x)^c$

$(A+x)\cap [-n+x,n+x]= (A\cap [-n,n])+x$


问:
Using induction to verify a statement
I have to prove that this statement is true.

For $n = 1, 2, 3, ...,$ we have $ 1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6$

Basically I thought I'd use induction to prove this. Setting $n = p+1$, I got this so far:

Left hand side: $1² + 2² + 3² +...+ p² + (p+1)² = p² + (p+1)² = 2p² + 2p + 1$

Right hand side: $(p + 1)(p + 1 + 1)(2 * (p + 1) +1)/6$

This is all I've got. I've never worked with induction so I don't even know if this correct to begin. Any insight or solution to my problem is appreciated.

答:
For proving an expression through induction. Check for the base case $n=1$,

Let $P(n)$ is true, $\sum_1^nk^2=1+4+9+ \dots n^2= P(n)=\dfrac{n(n+1)(2n+1)}{6}$.

You need to prove that $\sum_1^{n+1} (k)^2=1+4+9+ \dots (n+1)^2=P(n+1)=\dfrac{{(n+1)}{}(n+2)(2n+3)}{6}.$

Hint: $(k+1)^2=k^2+2k+1$ and $\sum_1^n k=\dfrac{n(n+1)}{2}, \sum_1^n 1=n$

Title: Using induction to verify a statement


答:
The Wikipedia article on Simpson's rule also gives your deviation, but suppresses the calculations that you could not master. Maybe you forgot to input that $x_0=a$, $x_1={1\over2}(a+b)$, $x_2=b$.

At any rate I would suggest to use another approach. We may assume $x_0=-1$, $x_1=0$, $x_2=1$ and verify easily that Simpson's rule gives the correct integral values for the three parabolas $y=ax^2+bx+c$ having $y=1$ at one of the three points and $y=0$ at the  other two points. This and the "linearity of everything" proves your claim.

Title: Derivation of Simspon's 1/3 rule using lagrange polynomial

Body:
Taking the Lagrange interpolation formular for a second order polynomial
$$f_2(x)=L_0(x)f(x_0)+L_1(x)f(x_1)+L_2(x)f(x_2)$$
Where $$\begin{align}L_0(x)&=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\\L_1(x)&=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}\\L_2(x)&=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\end{align}$$
Therefore,
$$f_2(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f(x_0)+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f(x_1)+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f(x_2)$$
The area of $f(x)$ under the interval $[x_0,x_2]$ is given by
$$\begin{align}\int_{x_0}^{x_2}f(x)dx&\approx\int_{x_0}^{x_2}\left[L_0(x)f(x_0)+L_1(x)f(x_1)+L_2(x)f(x_2)\right]dx\\\int_{x_0}^{x_2}f(x)dx&\approx\left[\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f(x_0)+\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f(x_1)+\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f(x_2)\right]dx\end{align}$$

After integrating I reached this$$\int_{x_0}^{x_2}f(x)dx\approx\left[\frac{\frac{x^3}{3}-\frac{x^2(x_2+x_1)}{2}+(x_1x_2)x}{(x_0-x_1)(x_0-x_2)}f(x_0)+\frac{\frac{x^3}{3}-\frac{x^2(x_2+x_0)}{2}+(x_0x_2)x}{(x_1-x_0)(x_1-x_2)}f(x_1)+\frac{\frac{x^3}{3}-\frac{x^2(x_0+x_1)}{2}+(x_1x_0)x}{(x_2-x_1)(x_2-x_0)}f(x_2)\right]$$ but after this I am unable to get the answer.
Can anyone help me, please?


问:
Dimension of normalizer of closed connected subgroup

Let $G$ be unipotent and let $H$ be proper closed, connected subgroup. Show that $\operatorname{dim}(N_G(H)) > \operatorname{dim} H$.

We know that $H \triangleleft Z_G(H) \triangleleft N_G(H)$, therefore I want to prove that one of the inclusion is proper.

I think the latter inclusion is proper, but I have no idea how to prove.

Any help will be appreciate..

答: Let $L_G, L_H,L_N$ be the respective Lie algebras of $G, H$ and $N_G(H)$. Consider the action of $H$ on $L_G/L_H$ by $u_x(y)=p([x,y])$ where $p:L_G\rightarrow L_G/L_H$ is the quotient map, the theorem of Engel implies the existence of $x_0\neq 0\in L_G$ such that $u_x(x_0)=0$ for every $x\in L_H$, we deduce that $x_0\in L_N$ and $dim(N_G(H))>dim(H)$.


答:
Since Tavish has already given the crucial key in a comment, I just add another way using the differentiation rule for inverse functions:

Set $$f(x) = \cosh(\sinh^{-1}x)$$

Then
$$f'(x) = \frac x{f(x)} \Rightarrow \left([f(x)]^2\right)' = 2x$$

Hence,

$$[f(x)]^2 = x^2+c \stackrel{f>0, f(0)=1}{\Longrightarrow}f(x) =\sqrt{x^2+1}$$

Title: Hyperbolic functions simplifying


问:
Find Ext's - 'easy' cases.
First exercise:

Let $k$ be a field and $R=k[x]/x^2$. Find $\operatorname{Ext}^{1}_{R}(k,k)$.

Second exercise:

Let $A$ be a torsion $\mathbb{Z}$-module. Prove $\operatorname{Ext   }^{1}_{\mathbb{Z}}(\mathbb{Z},A) = \operatorname{Hom}_\mathbb{Z}(A,U(1))$ where $U(1)$ is a multiplicative group of complex numbers of modulus $1$.

That domain of mathematics is crazy. I have no power here. I just want to survive first course of that and not continue learning those things.

答:
Exercise 1

As pointed out in the comments, $k$ has a free resolution as

$$\cdots \to R \xrightarrow{x} R \xrightarrow{x} R \to k \to 0$$

where $R \to k$ by $a + bx \mapsto a$ (equivalently, $x \mapsto 0$). Taking the deleted resolution and applying $\operatorname{Hom}_R(-, k)$ we get

$$0 \to \operatorname{Hom}_R(R,k) \xrightarrow{x^*} \operatorname{Hom}_R(R,k) \xrightarrow{x^*} \operatorname{Hom}_R(R,k) \to \cdots$$

where $x^*: \operatorname{Hom}_R(R,k) \to  \operatorname{Hom}_R(R,k)$ is the map induced by multiplication by $x$. This sends a map $\varphi: R \to k$ to the composition $R \xrightarrow{x} R \xrightarrow{\varphi} k$. Even more explicitly, we have $x^*(a+bx) = \varphi(x\cdot(a+bx)) = \varphi(ax) = ax\varphi(1)$. This is zero because the $R$-module structure on $k$ is given by $(ax+b)\cdot c = bc$.

Since $x^*$ is the zero map, the homology at position $1$ in the above complex is precisely $\operatorname{Hom}_R(R,k)$. Thus $\operatorname{Ext}_R^1(k,k) \cong \operatorname{Hom}_R(R,k)$. The latter is isomorphic to $k$ via $\varphi \mapsto \varphi(1)$, so $\operatorname{Ext}_R^1(k,k) \cong k$.

Exercise 2

The exercise as stated is incorrect, since $\operatorname{Ext}_\mathbb{Z}^1(\mathbb{Z},M) = 0$ for any abelian group $M$ since $\mathbb{Z}$ is projective. The correct statement is $\operatorname{Ext}_\mathbb{Z}^1(A, \mathbb{Z}) \cong \operatorname{Hom}_\mathbb{Z}(A, U(1))$.

For this, consider the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to U(1) \to 0$ where the first map is inclusion and the second map sends $\theta$ to $e^{i \theta}$. This gives a long exact sequence for $\operatorname{Ext}_\mathbb{Z}^1(A, -)$:

$0 \to \operatorname{Hom}_\mathbb{Z}(A, \mathbb{Z}) \to \operatorname{Hom}_\mathbb{Z}(A, \mathbb{R}) \to \operatorname{Hom}_\mathbb{Z}(A, U(1)) \to \operatorname{Ext}_\mathbb{Z}^1(A, \mathbb{Z}) \to \operatorname{Ext}_\mathbb{Z}^1(A, \mathbb{R}) \to \cdots$

However, note that $\operatorname{Hom}_\mathbb{Z}(A, \mathbb{R})=0$ since $A$ is torsion: for any element $a \in A$, a homomorphism $A \to \mathbb{R}$ must send $a$ to a torsion element in $\mathbb{R}$, which is torsion-free. Lastly, $\operatorname{Ext}_\mathbb{Z}^1(A, \mathbb{R})$ is zero since $\mathbb{R}$ is injective (it's a divisible abelian group).

Therefore, the map in the long exact sequence $\operatorname{Hom}_\mathbb{Z}(A, U(1)) \to \operatorname{Ext}_\mathbb{Z}^1(A, \mathbb{Z})$ is an isomorphism.

Title: Find Ext's - 'easy' cases.


问:
How many ways? - difficult combinatorics
These are from an archival local math contest:

The city purchased train rails in two colors: gray and black. The only difference between them is the color. The project assumes that the rails will be installed in series of 10. We say that the series is elegant when no two black rails are directly adjacent to each other. How many different elegant series of 10 rails are possible?

I made some observations:

In 3. I know the answer is supposed to be 144. If we have 6 black rails or more we cannot make an elegant series. For 1 black rail we can make 10 different series (the order matters), for two it is 8 + 7 + ... + 1 = 36 (i placed the first rail and looked where I can place the second one). For 5 rails there are two elegant series. However I cannot calculate the numbers for 3 or 4 black rails, is there any nice and clean way?

答: A good way to count 3's and 4's is to "fix" the first rail, and the second, variate the third, then variate the second and so on. You could try complementary counting, but to my knowledge, casework is the best approach.


问:
Evaluating order of convergence
I think this is quite a simple question, I just want to make sure I understood all correctly.

Here's the problem: I have a numerical method, which is in some way dependent on its spacing $h$ (like pretty much every numerical method ;)).

Now I have calculated the errors of this method for different values of $h$, and I'm interested in the order of convergence.

What I've got so far: My current results always divide the spacing by 2, so $h_1 = \frac{h_0}{2}$, $h_2 = \frac{h_1}{2}$ etc.
I also have for the corresponding errors: $e_0 \cdot C = e_1$, $e_1 \cdot C = e_2$ with $C < 1$ and it is approx. constant for all steps.

So I guess in this case I have linear convergence $\mathcal{O}(h)$, right?

But my question is now: What if I don't have this relationship $h_i = \frac{h_{i-1}}{2}$, but instead use arbitrary spacings, say $h = \frac 1 2, \frac 1 3, \frac 1 4, ...$

What is my approach then to calculate the order of convergence?
Thanks

答:
Plot a graph of log(error) vs log(h). The slope of the curve gives the convergence rate. This can be easily computed using octave:

```<code>r=polyfit(log(h),log(error),1);
convergence_rate = r(1);
</code>```

Title: Evaluating order of convergence

Body:
Plot a graph of log(error) vs log(h). The slope of the curve gives the convergence rate. This can be easily computed using octave:

r=polyfit(log(h),log(error),1);
convergence_rate = r(1);


问:
Grothendieck connections and jets
The following question is based on some remarks in section I.2 of Deligne's book Equations Différentielles à Points Singuliers Réguliers.

Let $X$ be a smooth complex variety and $X_1$ the first infinitesimal neighborhood of the diagonal in $X \times X$, so there is a natural morphism $X \to X_1$. If we write $p_1,p_2 : X_1 \to X$ for the two projections, then the first-order jet bundle of a vector bundle $V$ on $X$ is defined to be $J^1(V) = p_{1*} p_2^*V$. Here "upper star" is used in the sense of $\mathcal{O}$-modules. This allows for a convenient way of expressing the notion of a connection: this is just an isomorphism $p_1^*V \to p_2^*V$ which restricts to the identity over $X$, which is the same as an $\mathcal{O}$-linear map $V \to J^1(V)$ such that the composition $V \to J^1(V) \to V$ is the identity.

Here Deligne says something I don't understand: he refers to a first-order differential operator $j^1 : V \to J^1(V)$ "which associates with any section a first-order jet" (I'm translating from the French). What is he talking about? I don't have much intuition for jets, although I know they have something to do with taking Taylor expansions, so any general explanation of that would be greatly appreciated as well.

答:
Question: "Thanks for the helpful explanation of the connection with Taylor expansions. I'm still unsatisfied with the definition of j1, partially because this "pointwise definition" of the jet bundle can't be the whole story. After all, one can't in general construct a sheaf just by specifying its stalks. There must be a way to say it with the global definition I gave above: more explicitly, J1(V)=p2∗(OX1⊗OXp−12V)."

Answer. There is an elementary discussion of jets and Taylor expansions of rational functions at the following link:

https://mathoverflow.net/questions/55244/why-must-nilpotent-elements-be-allowed-in-modern-algebraic-geometry/378811#378811

In general if $A\rightarrow B$ is a map of commutative unital rings and if $E$ is any $B$-module, you may define the $l$'th jet bundle $J^l(E):=B\otimes_A B/I^{l+1}\otimes_B E$. There is a canonical map

T1. $T^l: E\rightarrow J^l(E)$

defined by $T^l(e):=1\otimes 1\otimes e$. When $l=1$ you get the map Deligne speaks of in his book. There is when $J^1$ is a finite rank projective $B$-module and $E$ is a finite rank projective $B$-module, a short exact sequence of left $B$-modules

T2. $0 \rightarrow \Omega^1_{B/A}\otimes_B E \rightarrow J^1(E) \rightarrow E \rightarrow 0,$

and the sequence T2 is split iff there is a connection $\nabla: E \rightarrow \Omega^1_{B/A}\otimes_B E$. This gives a "global" definition of the Taylor morphism $T^l$. The Taylor morphism is a differential operator of order $\leq l$:

T3. $T^l\in \operatorname{Diff}^l_A(E, J^l(E))$

for any $l \geq 1$.
If $A:=k$ is a field and $\mathfrak{m}\subseteq B$ corresponds to a $k$-rational point, it follows there is an isomorphism

T4.  $J^l(\mathfrak{m}):= J^l\otimes_{B_{\mathfrak{m}}}\kappa(\mathfrak{m}) \cong  B/\mathfrak{m}^{l+1}$.

More generally $J^l(E)(\mathfrak{m})\cong E/\mathfrak{m}^{l+1}E$, and the Taylor morphism $T^l(\mathfrak{m})$ is the canonical map

T5. $T^l(\mathfrak{m}): E \rightarrow E/\mathfrak{m}^{l+1}E$

defined by $T^l(\mathfrak{m})(e):=\overline{e}$.

Example. Let $M$ be a complex manifold with structure sheaf $\mathcal{O}_M$ and let $\mathcal{E}$ be a locally trivial $\mathcal{O}_M$-module of finite rank. Let $\mathcal{O}_{M\times M}$ be the structure sheaf of the product manifold and let $p,q:M\times M \rightarrow M$ be the two projection maps. You may define $J^l_M:=\mathcal{O}_{M\times M}/I^{l+1}$ where $I$ is the "ideal of the diagonal", and $J^l_M(\mathcal{E}):=p_*(J^l_M \otimes q^*\mathcal{E})$.
There is a Taylor morphism,

T6. $T^l: \mathcal{E}\rightarrow J^l_M(\mathcal{E})$

similar to the map T1 defined in the algebraic case. Note that when $f:X \rightarrow Y$ is a map of complex manifolds, the pull back $f^*\mathcal{E}$ is defined as follows. The map $f$ induce a map

T7. $f^{\#}:\mathcal{O}_Y \rightarrow f_*\mathcal{O}_X$.

Let $U\subseteq Y$ be an open set and let $f_U:f^{-1}(U)\rightarrow U$ be the restricted map. Since $f$ is holomorphic it follows $f_U$ is holomorphic.
If $s\in \mathcal{O}_Y(U)$ is a holomorphic function it follows $s\circ f\in \mathcal{O}_X(f^{-1}(U))$ is a holomorphic function, inducing the map $f^{\#}$. We get an induced map $\tilde{f}: f^{-1}(\mathcal{O}_Y)\rightarrow \mathcal{O}_X$. Since $\mathcal{E}$ is an $\mathcal{O}_Y$-module it follows $f^{-1}(\mathcal{E})$ is an $f^{-1}(\mathcal{O}_Y)$-module, and we define

T8. $f^*\mathcal{E}:=\mathcal{O}_X\otimes_{f^{-1}(\mathcal{O}_Y)}f^{-1}(\mathcal{E})$.

It follows the left $\mathcal{O}_X$-module $J^l(\mathcal{E})$ is a locally trivial sheaf of finite rank on $X$. I believe Hartshornes Exercise II.5.18 is true in this case, hence there is an equivalence of categories between the category of finite rank locally trivial sheaves on $X$ and the category of finite rank holomorphic vector bundles on $X$. Hence to $J^l(\mathcal{E})$ you should get a holomorphic vector bundle $J^l_{h}(\mathcal{E})$ whose fiber is the fiber described above. This gives a global definition valid for any finite rank  locally free sheaf on any complex manifold.

Title: Grothendieck connections and jets


问:
What is "box-constrained mathematical optimization problem"?
I came across the term "box-constrained mathematical optimization problem" while reading a research paper.

Can someone please explain what 'box-constrained' and "box-constrained mathematical optimization problem" mean ?

答:
It is a problem where the only constraints are upper and lower bounds on the variables (which can be $\pm \infty$). Examples are:
$$\min_{x \geq 0} f(x)$$
$$\min_x \{ f(x) : 0 \leq x_i \leq 1 \; \forall i\}$$
$$\min_x \{ f(x) : l_i \leq x_i \leq u_i \; \forall i\}$$


问:
Show that $r^n\rightarrow 0$ if $|r|<1$
Let $r\in \mathbb{C}$, $|r|<1$. Show that $r^n\rightarrow 0$.

For that we have to show that $r^n$ is decreasing and bounded, right?

Let $a_n:=r^n$.

Then we have that $\frac{a_{n+1}}{a_n}=\frac{r^{n+1}}{r^n}=r\leq |r|<1$ and so the sequence is decreasing.

Since $|r|<1$ we get that $-1<r<1\Rightarrow (-1)^n<r^n<1^n \Rightarrow (-1)^n<r^n<1$ and so the sequence is bounded.

So we know that the sequence converges to a limit $L$.

We have that $r^{n+1}=r\cdot r^n$. Since $r^n\rightarrow L$ we also have that $r^{n+1}\rightarrow L$ and so we get for $n\rightarrow \infty$ that $L=r\cdot L\Rightarrow L\cdot (r-1)=0$ and since $|r|<1$ and so $r\neq 1$ it follows that $L=0$.

Is everything correct?

答:
No, it is not. Even assuming that $r\in\Bbb R$, you did not prove correctly that the sequence is decreasing, and you could not have possibly have done so, since, if $-1<r<0$, the sequence is not decreasing (nor increasing, for that matter).

Besides, you don't have to prove that the sequence is bounded and decreasing.

Title: Show that $r^n\rightarrow 0$ if $|r|<1$


问:
Linear Least Squares with Linear Equality Constraints: $ \arg \min_{x} {\left\| x - {x}_{0} \right\|}_{2}^{2} $ s. t. $ A x = b $
Let $A\in \mathbb R^{m×n}, b \in \mathbb R^m$ with $m \leq n$, $\operatorname{rank} A = m$ and $x_0 \in \mathbb R^n$. Consider the problem :
$$\arg \min\|x-x_0\|² \quad\text{s.t. }Ax = b$$

How can I solve this, please?
I was thinking about using the least square solution to $Ax=b$ minimizing $\|x\|$ but I'm not sure.

Thank you for your help.

答:
The general problem is given by (Linear Least Squares with Linear Equality Constraints):

$$
\begin{alignat*}{3}
\arg \min_{x} & \quad & \frac{1}{2} \left\| A x - b \right\|_{2}^{2} \\
\text{subject to} & \quad & C x = d
\end{alignat*}
$$

The Lagrangian is given by:

$$ L \left( x, \nu \right) = \frac{1}{2} \left\| A x - b \right\|_{2}^{2} + {\nu}^{T} \left( C x - d \right) $$

From KKT Conditions the optimal values of $ \hat{x}, \hat{\nu} $ obeys:

$$ \begin{bmatrix}
{A}^{T} A & {C}^{T} \\
C & 0
\end{bmatrix} \begin{bmatrix}
\hat{x} \\
\hat{\nu}
\end{bmatrix} = \begin{bmatrix}
{A}^{T} b \\
d
\end{bmatrix} $$

Now all needed is to solve the above with any Linear System Solver.

To adapt this into your problem we can set $ A = I $ and solve.


问:
In △ ABC, D is the midpoint of AB, while E lies on BC satisfying BE = 2EC. If m∠ADC=m∠BAE, what is the measure of ∠BAC in degrees?
In △ABC, D is the midpoint of AB, while E lies on BC satisfying BE = 2EC. If m∠ADC=m∠BAE, what is the measure of ∠BAC in degrees? I know already that angle A and angle D are congruent because m∠ADC=m∠BAE I just dont know how to approach the problem and i would love some help

答:
edit Given that this was an NCTM calendar problem, I doubt that my solution is the intended one, but I've added some more detail to flesh it out.

Label the point of intersection of $\overline{AE}$ and $\overline{CD}$ as $X$.

Use the technique of mass points:

Since $D$ is the midpoint of $\overline{AB}$, $AD:DB=1:1$.

Since $BE=2\cdot EC$, $BE:EC=1:2$.

Putting masses of $1$ at $A$, $1$ at $B$ and $2$ at $C$ will allow the triangle to balance at $X$.

The "effective mass" at $D$ is $2$ (the sum of the masses at $A$ and $B$).

The ratio $CX:XD$ is equal to the ratio of the masses $D:C$, so $CX:XD=2:2=1:1$, or $CX=XD$.

From $m\angle ADC=m\angle BAE$, $\triangle ADX$ is isosceles with $AX=DX$.

So we have $AX=DX=CX$.

So $A$, $C$, and $D$ are all the same distance from $X$.  A Circle is the set of points that are a fixed distance from a fixed point.  So there is a circle with center $X$ that contains $A$, $C$, and $D$.

$\overline{CD}$ is a diameter of the circle, since it contains the center $X$.

$\angle DAC$ is inscribed in the circle, or, since $\overline{CD}$ is a diameter, $\angle DAC$ is inscribed in a semicircle.  From this, you can conclude the measure of $\angle DAC$, which is the same as the measure of $\angle BAC$.


问:
integration by substitution, using $\;t = \tan \left(\frac 12 x\right)$
$\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2-\cos x} \, dx$ using the substitution $t=\tan\frac{1}{2}x$

$x=2\tan^{-1}t$

$\dfrac{dx}{dt}=\dfrac{2}{1+t^2}$

$dx=\dfrac{2}{1+t^2}\,dt$

$\displaystyle\int_0^1 \left(\frac{1}{2-\cos x}\right)\left(\frac{2}{1+t^2}\right)\,dt$

Is this the right idea? If so what do I do next?

$\displaystyle\int_0^1\left(\frac{1}{2-\frac{1-t^2}{1+t^2}}\right) \,\left(\frac{2}{1+t^2}\right)\, dt$

$\displaystyle\int_0^1\frac{2}{1+3t^2}\,dt$

$=2\left[\frac{\ln(1+3t^2)}{6t}\right]_0^1$

答:
If $x=2\arctan t$ then $\cos x = \cos\left(2\arctan t\right)$.  Use the fact that $\cos(2u)=\cos^2u-\sin^2u$.  So you get
$$
\cos x = \cos^2\left(\arctan t \right) -\sin^2\left(\arctan t \right)
$$

If $\varphi=\arctan t$ then $\dfrac t1=t = \tan\varphi=\dfrac{\text{opposite}}{\text{adjacent}}$, so $\cos\varphi=\dfrac{\text{adjacent}}{\text{hypotenuse}}$.  So $\text{opposite}=t$, $\text{adjacent}=1$, and by the Pythagorean theorem $\text{hypotenuse}=\sqrt{1+t^2}$.

Then $\cos=\dfrac{\text{adjacent}}{\text{hypotenuse}}$ and so
$$
\cos^2(2\arctan t) = \cos^2 (\cdot)-\sin^2(\cdot) = \left(\frac{1}{\sqrt{1+t^2}}\right)^2 - \left(\text{something similar}\right)^2 = \cdots\cdots
$$


答: So $|z^n - 0| = |z^n| = |z|^n \rightarrow 0$, which is the definition of $z^n \rightarrow 0$.

Title: Show that $r^n\rightarrow 0$ if $|r|<1$

Body:
Let $r\in \mathbb{C}$, $|r|<1$. Show that $r^n\rightarrow 0$.

For that we have to show that $r^n$ is decreasing and bounded, right?

Let $a_n:=r^n$.

Then we have that $\frac{a_{n+1}}{a_n}=\frac{r^{n+1}}{r^n}=r\leq |r|<1$ and so the sequence is decreasing.

Since $|r|<1$ we get that $-1<r<1\Rightarrow (-1)^n<r^n<1^n \Rightarrow (-1)^n<r^n<1$ and so the sequence is bounded.

So we know that the sequence converges to a limit $L$.

We have that $r^{n+1}=r\cdot r^n$. Since $r^n\rightarrow L$ we also have that $r^{n+1}\rightarrow L$ and so we get for $n\rightarrow \infty$ that $L=r\cdot L\Rightarrow L\cdot (r-1)=0$ and since $|r|<1$ and so $r\neq 1$ it follows that $L=0$.

Is everything correct?


答:
I will answer the third one first.

For 1,2 and 5 black rails you have already calculated it so I would not bother calculating again.

We will use the gap method

Case 1: 3 black rails

First, you lay down 7 grey rails so there are 8 gaps

Out of those 8 gaps, you select 3 because there are 3 black rails.

So $\dbinom{8}{3} = 56$

Case 2: 4 black rails

You lay down 6 grey rails so 7 gaps are left and you have to select 4 for 4 black rails

.
So $\dbinom{7}{4} = 35$

You have made an error in the case where there are 5 black rails. The answer should be 6 for that case(6C1)

Total = 10 + 36 + 56 + 35 + 6 +1= 144

Title: How many ways? - difficult combinatorics

Body:
These are from an archival local math contest:

The city purchased train rails in two colors: gray and black. The only difference between them is the color. The project assumes that the rails will be installed in series of 10. We say that the series is elegant when no two black rails are directly adjacent to each other. How many different elegant series of 10 rails are possible?

I made some observations:

In 3. I know the answer is supposed to be 144. If we have 6 black rails or more we cannot make an elegant series. For 1 black rail we can make 10 different series (the order matters), for two it is 8 + 7 + ... + 1 = 36 (i placed the first rail and looked where I can place the second one). For 5 rails there are two elegant series. However I cannot calculate the numbers for 3 or 4 black rails, is there any nice and clean way?


问:
Maximum Modulus theorem applied on mapping
Question: For $|z_0|<R$, I want to show that the mapping $$T(z)=\frac {R(z-z_0)} {R^2-\bar{z_0}z}$$ takes the open disc of radius $R$ $1-1$ and onto the unit disc and $z_0\rightarrow 0$.

Hint: Use the max modulus theorem and verify that $z_0\rightarrow0$ and $|z|=R\implies |T(z)|=1$.

this is how the question was posed with the hint.

Approach: I covered the automorphism $\phi_\alpha(z)=\frac{z-\alpha}{1-\bar{\alpha}z}$ of unit disc mapping onto a unit disc in class. But since $|z_0|<R$ inside the disc of radius R, I took $\frac{|z_0|}{R}<1$ and plugged it in $\phi$ to get $\phi_{\frac{z_0}{R}}=\frac{Rz-z_0}{R-\bar z_0z}$. Which is not what i want.

Also i tried the straight forward approach to show its 1-1 and onto, but the onto part looks like this: $$z=\frac{R^2w+Rz_0}{w\bar z_0+R}$$
Is this right?
Also I'm so confused as to how to use the max modulus theorem in this case.

Any help would be greatly appreciated.

Thankyou!

答:
d13: If you had divided both $z_0$ and $z$ by $R$, you'd map the disk of radius $R$ to the disk of radius $1$: Letting $\alpha = z_0/R$ and $Z=z/R$,
$$T(z) = \frac{R(z-z_0)}{R^2-\bar z_0z} = \frac{R^2(Z-\alpha)}{R^2(1-\bar\alpha z)} = \frac{Z-\alpha}{1-\bar\alpha Z}\,,$$
with $|Z|<1$. Now we apply the standard knowledge about the automorphisms of the unit disk.


问:
Convergence of sequences in the strong Whitney topology
Let $\{f_n\}_{n \in \mathbb{N}}$ a sequence of $C^r(M, N)$ functions of paracompact manifolds converging to $g$ in the strong Whitney topology. Prove that there exist a compact $K$ of $M$ $f_n$ and $g$ agree every where except on $K$, $\forall n \in \mathbb{N}$

I do not see how to proceed as the strong Whitney topology (as defined in Hirsh' Differential Topology) only guarantees that the $f_n$ (and their derivatives) are $\epsilon_i$-near of $g$ on compacts $K_i$.

Thank you in advance.

答:
To prove this we will go by contradiction, and our main goal now is to construct an open set of $C_S^r(M,N)$ that will not contain an infinite number of $f_n's$,and so contradicting the fact that $f_n\rightarrow g$ in $C_S^r(M,N)$.For this we need to take a compact exhaustion of $M$, that is compact sets $\{K_i\}_{i=1}^{\infty}$ such that their interiors cover $M$ and $K_1\subset int(K_2)\subset K_2 \subset ...$ \.
First we show that there exists a sequence of $x_k's$ in $M$ , and increasing sequences $n_k$ and $l_k$, such that $x_k\in K_{l_k}-int(K_{l_{k-1}})$ and $f_{n_k}(x_k)\neq g(x_k)$. This is done by induction . We have that $K_0= \emptyset$ and take $N=1$, and we have that there exists $x_1\in M$, more precisely in some $K_i$ which we now denote $K_{l_1}$, such that $f_{n_1}(x_1)\neq g(x_1)$, where $n_1>N$.Now suppose we have found $x_1,...x_{k-1}$, $n_{k_1},...,n_{k-1}$ and $l_{k_1},...,l_{k-1}$.Now set $N=n_{k-1}+1$ and consider $K=K_{l_{k-1}}+1$. We have that there exists $x_k\in M-K$ and $n_{k}>N$ such that $f_{n_k}(x_{k})\neq g(x_k)$. We have that $x_k$ belongs to some $K_i$ and so we consider $l_k=i$ and we have that $x_k\in K_{l_k}-int(K_{l_{k-1}})$.\
Now that we have the previous property we now construct the desired open set. First notice that each $x_k\in int(K_{l_k+1})-K_{l_k-2}$, and that these will be open sets.Now consider a coordinate chart around each $x_k$, $(U_k,\phi_k)$, and by doing intersections if needed, we can assume that we have compact sets $K'_k$ and locally finite charts $(U_k,\phi_k)$ such that $K'_k\subset U_k\subset int(K_{l_k+1})-K_{l_k-2}$, furthermore we can assume that there exists charts $(V_k,\psi_k)$ for $g(x_k)$ such that $g(U_k)\subset V_k$. Now since we have that $f_{n_k}(x_k)\neq g(x_k)$ we have that there exists $\epsilon_k$ such that $||\psi_k\circ f_{n_k}(x_k)-\psi_k\circ g(x_k)||>\epsilon_k$. Now putting all of this together consider $\Phi=\{U_k,\psi_k\}_{k=1}^{\infty}, \Psi=\{V_k,\phi_k\}_{k=1}^{\infty}, K=\{K_k\}_{k=1}^{\infty}$ and $\epsilon=\{\epsilon_k\}_{k=1}^{\infty}$, we will have that $N^r(g;\Phi,\Psi,K,\epsilon)$ will be our desired neighborhood.


答:
As for any complex number $$|r^n-0|=||r|^n-0|$$ it suffices to show that the property holds in the reals.

Let $s:=|r|>0$. Now for any $\epsilon>0$, $|s^n-0|<\epsilon$ if $n\ge\log_s\epsilon$, and the latter expression is always defined. (Needless to say, $s=0$ also works.)

Title: Show that $r^n\rightarrow 0$ if $|r|<1$


问:
Quasi-linear PDE $(1-xy)u_x+u_y+u_z=(y-z)u$ in 3 variables
How do I go about solving $(1-xy)u_x+u_y+u_z=(y-z)u$? I know how to solve quasi-linear PDEs with 2 independent variables, but solving with three is giving me problems.

So me and my friend are trying to self learn PDEs. We're not sure how to do it without boundary conditions. (How is it different?) But with a boundary condition, say $u(y,0)=f(y)$ for some function $f(y)$, let $y(0)=s$, $z(0)=0$, and $w(0)=f(y)$, then we have $dy/dt=1⟹y=t+s$, $dz/dt=1⟹z=t$, and $dw/dt=(y−z)w=sw⟹w=f(y)e^{st}$, so $u(y,z)=f(y)e^{(y−z)z}$. This is quite the problem we run into when trying to solve the 3-variable PDE I posted; how does $u(x,y,z)$ depend on $x$?

答:
$$(1-xy)u_x+u_y+u_z=(y-z)u$$
https://en.wikipedia.org/wiki/Method_of_characteristics

Charpit-Lagrange characteristic ODEs :
$$ds=\frac{dx}{1-xy}=\frac{dy}{1}=\frac{dz}{1}=\frac{du}{(y-z)u}$$
A first characteristic equation comes from solving $\frac{dy}{1}=\frac{dz}{1}$ :
$$y-z=c_1$$

A second characteristic equation comes from solving $\frac{dx}{1-xy}=\frac{dy}{1}$ :
$$x \: e^{\frac12 y^2}-\sqrt{\frac{\pi}{2}} \: \text{erfi}\left(\frac{y}{\sqrt{2}}\right)=c_2$$
This is not a simple function, but the equation of a characteristic curve involving a special function namely erfi : https://mathworld.wolfram.com/Erfi.html

A third characteristic equation comes from solving $\frac{dy}{1}=\frac{dz}{1}=\frac{ydy-zdz}{y-z}=\frac{du}{(y-z)u}$ :
$$e^{-\frac12(y^2-z^2)}u=c_3$$

The general solution of the PDE expressed on the form of implicit equation
is $\Phi\left(c_1,c_2,c_3\right)=0$ where $\Phi$ is an arbitrary function of three variables. Or explicitly $u=e^{\frac12(y^2-z^2)}F(c_1,c_2)$ :
$$\boxed{u(x,y,z)=e^{\frac12(y^2-z^2)}F\left((y-z)\:\: , \:\:x \: e^{\frac12 y^2}-\sqrt{\frac{\pi}{2}} \: \text{erfi}\left(\frac{y}{\sqrt{2}}\right)\right)}$$
where $F$ is an arbitrary function of two variables.

The function $F$ has to be determined if some boundary conditions are specified.


问:
Proper inclusion of quasi-analytic classes
It is known that $C\{n!\}$ is a quasi-analytic class. I was asked to find a quasi-analytic class $C\{M_n\}$ such that $C\{n!\}$ is a proper subclass of $C\{M_n\}$. My though is that $C\{n!\} \subseteq C\{(2n)^n\}$ by definition. By the Denjoy-Carleman theorem, we see that $C\{(2n)^n\}$ is also quasi-analytic. However, I can't find a function $f \in C\{(2n)^n\}$ but $f \notin C\{n!\}$. Any suggestion?

答:
Hint:

Note that $f_n(x)=f(nx)$ where $f=f_1$ is quasi-analytic of class $n!$ since it is bounded on the reals and analytically extendable into a strip; since $(f_n)^{(k)}(x)=n^kf^{(k)}(nx)$ one gets a bound for $g(x)=\sum a_n f_n(x), a_n >0$ of the type $|g^{(k)}(x)| \le CA^k k!\sum a_n n^k =CA^k k! C_k$ and by Denjoy Carleman one needs $\sum \frac{C_{k-1}}{kC_k} = \infty$ so one needs $C_k$ to grow not that fast, but more like $(\log k)^k$ which insures $g$ is quasi-analytic

($C_k$ can get quite big for large $k$ regardless of how small $a_n$ is, since one can fix an $N$ and pick $N^k \to \infty, k \to \infty$ much bigger than $a_N$ - also obviously $g$ is not extendable to any strip but it still bounded if $\sum a_n$ finite so $g$ is not of class $k!$)

But using the fact that now if the sum converges for all $k$ and $C_k$ is finite as it must be of course, then $a_nn^k \to 0, n \to \infty, k$  fixed, so $C_k$ is about like its largest term for about an interval of the same magnitude in $n$ around said largest term, at least if $a_n$ is fast decreasing, one actually needs that the largest term to be where $n$ is about $\log k$ (or less) so $a_n$ for example can be taken as $1/n^{n^n}$.

This ensures that for fixed $k$, $a_nn^k$ attains its plateau for $n^n$ close to $k$ and decreases very fast once $n^n$ gets much bigger than $O(k)$. With this choice fixing $N=N_k$ st $N^N$ is closest such to $k$, we get $N \log N$ about $\log k$ so the right order of magnitude we want for $C_k$ as then we can estimate $a_n=\frac{n^{N^N}}{n^{N^N}} \le N^{N^N}, n \le N$ so $a_n \le (\log k)^{k}$ for some $N=O(\log k)$ values and much lower thereafter, giving the right estimate for $C_k$ by $O(\log k)(\log k)^{k}$. Conversely similar estimates show that $C_k \ge (1+B_k)(\log k)^{k}, B_k$ asymtpotically negligible, so the ratio $C-{k-1}{C-k}$ grows like $1/\log k$ up to asymptotic dactors $1+B_k, B_k$ negligible

This last part needs some clear estimates but I think they are doable with the hint provided.


问:
Relations on Sets
Let $A \neq \emptyset$ and $O$ an antisymmetric relation on $A$. Suppose that for every $B \subseteq A$, $B \neq \emptyset$, $\exists$ least $B$ i.e. $\exists b \in B$ s.t $\forall c \in B$ $bOc$. Now I want to proof that $O$ is a total order in $A$. I haven't really made any progress in this question, I know that for being a total order, a relation $R$ has to be reflexive, transitive, antisymmetric and total. I'm given $O$ is antisymmetric, but from the least element existence I don't know how to proceed.

答:
Take $x,y\in A$. Then by considering $B=\{x,y\}$ and applying the hypothetis, we have that $x$ or $y$ is a minimum. Let's suppose it's $x$. Then $xOy$. And if it's $y$, then $yOx$. So the relation is total.

Take $x\in A$. Then by considering $B=\{x\}$ we have that $x$ is the miminum : $xOx$. So the relation is reflexive.

Take $x,y,z\in A$ such that $xOy$ and $yOz$. By considering $B=\{x,y,z\}$, we have that the minimum is $x,y$ or $z$. It can't be $z$ since $yOz$, and it can't be $y$ since $xOy$. So it's $x$. So $xOz$.

Title: Relations on Sets


问:
When we write $y'' + y' + y = x$, for a function $y$, do we really mean $y''(x) + y'(x) + y(x) = x$?
I have recently started exploring differential equations at a graduate level (partial and ordinary) and really started thinking about why we write them the way we do. Let (since this seems to be the common symbol of choice) $y: \mathbb{R} \to \mathbb{R}$ such that $x \mapsto y(x)$. Just to be fussy, we will consider $y \in C^{\infty}(\mathbb{R})$. Now, for example, consider the second order differential equation $$y'' + y' + y = x \tag{i}$$ My question is in regards to this notation: Why do we write the function and it's derivatives this way in the equation rather than writing: $$y''(x) + y'(x) + y(x) = x \tag{ii}$$ using the function values? When we write (i), do we really mean (ii)? This may be a bit pedantic, but I just want to be thorough.

答:
The context of the equation
\begin{equation}
y^{\prime\prime} + y^{\prime} + y = x
\end{equation}
means that given a function described by the following proposition:
\begin{equation}
\forall x \in \mathbb{R}, y^{\prime\prime}\left(x\right) + y^{\prime}\left(x\right) + y\left(x\right) = x,
\end{equation}
use deduction rules to deduce a new proposition in the following form:
\begin{equation}
\forall x \in \mathbb{R}, y\left(x\right) = ...
\end{equation}
Of course, rigorously speaking, you should first prove the existence and uniqueness of such a function $y$:
\begin{equation}
\exists! y: \mathbb{R} \mapsto \mathbb{R}, \forall x \in \mathbb{R}, y^{\prime\prime}\left(x\right) + y^{\prime}\left(x\right) + y\left(x\right) = x,
\end{equation}
which cannot be the case without boundary values. Thus, to specify a unique solution to your equation, you need to provide boundary values.

In the end, you conclude the following proposition as a solution to the problem:
\begin{equation}
\forall y:\mathbb{R} \mapsto \mathbb{R}, \left[\forall x \in \mathbb{R}, y^{\prime\prime}\left(x\right) + y^{\prime}\left(x\right) + y\left(x\right) = x \wedge y\left(0\right) = ... \wedge y^{\prime}\left(0\right) = ...\right] \longrightarrow \left[y\left(x\right) = ...\right]
\end{equation}

It should be noted that providing an analytical solution is not the only way to solve a differential equation (ordinary or partial). From set theory, functions are sets of 2-tuples. As long as you have a way to specify each 2-tuple in a function, the differential equation is solved. In this context, numerical approaches are also a convenient way to specify 2-tuples of a function.

A lot of times, people don't think about the logical meaning behind solving equations. Education of logic systems in math is often ignored in today's education system. Recently, someone even told me training in first order logic is not important to mathematicians. This is exactly the reason why math is so formidable to most people: using daily language to talk about a thing in the world of object language is not gonna lead to the success of most people. People fear math as math is not reachable to them. To make people truly understand the tool of math (they don't have to be masters using this tool for creative inventions), languages should be taught again from scratch.


问:
Eigenfunctions of $L = \nabla^2 + \delta(x-x_o)$
I am looking for the eigenfunctions and eigenvalues of

$L = \nabla^2 + \delta(x-x_0)$, $\quad x \in \mathbb{R}^n, \quad n = 1, 2.$

The idea is to solve $(\nabla^2 + \delta(x-x_0))p(x) = -k^2p(x)$ $\quad(1).$

I have found nothing similar to this in the literature, any help is more than welcome.

答:
Since the equation is translation invariant, we may set $x_0=0$ with no loss of generality. When $n=1$, the eigenproblem is

$$\tag{1}
p''(x)+\delta(x)p(x)=-k^2p(x)
$$

Integrating (1) over a vanishing region around $x=0$ leads to the derivative discontinuity condition

$$\tag{2}
p'(0^+)-p'(0^-)+p(0)=0
$$

Continuity of $p$ at $x=0$ requires $p(0^+)=p(0^-)$. A direct approach would be to solve (1) in the regions $x<0$ and $x>0$ then patch the solutions together using (2) and the continuity condition. Instead, let us make an ansatz for the eigenfunctions of (1) by noting that for $x\neq0$ the solutions are sinusoids and that we can consider the even $P_k$ and odd $Q_k$ solutions separately

$$\tag{3}
P_k(x)=\sin(k|x|+\alpha_k)\\
Q_k(x)=\sin(kx+\beta_k)
$$

Substituting (3) into (1)  we find the constants

$$\tag{4}
\alpha_k=\tan^{-1}(-1/2k)\\
\beta_k=0
$$

When $n\geq 2$, it is not so simple. One needs to work with a regularized version of the delta (as done here). For example, with $n=2$ in polar $(r,\phi)$ co-ordinates, one choice is

$$\tag{5}
\delta_R(r)=\frac{H(R-r)}{\pi R^2}
$$

Then we would need to solve

$$\tag{6}
\frac{\partial^2 p}{\partial r^2}+\frac{1}{r}\frac{\partial p}{\partial r} + \frac{\partial^2 p}{\partial \phi^2} + \delta_R(r) p =-k^2 p
$$

Then, at the end of the calculation, we study the solutions as $R\to 0$. Since this is done in the linked paper, I will briefly describe a method:

Separate variables $p(r,\phi)=f(r)g(\phi)$, substitute into (6) and get two ODEs.

The ODE for $g(\phi)$ will be simple and have solutions $g(\phi)=e^{im\phi}$ with integer $m$.

The ODE for $f(r)$ needs to be further split into regions $r<R$ and $r>R$. Call these solutions $f_-$ and $f_+$. They will be Bessel functions.

Enforce continuity of $f$ and its derivative at $r=R$ so that $f_-(R)=f_+(R)$ and $\partial_r f_-(R)=\partial_r f_+(R)$.